Let \((a,n) \in \hspace{0.05em} \mathbb{N}^2\) be two natural numbers.
The sum of natural numbers\( : \sum k\)
The sum of the \( (n + 1) \) first natural numbers, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k = 1 + 2 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (n-1) + n \hspace{0.1em} \hspace{0.1em} = \frac{n(n+1)}{2} $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k = \frac{(n + a)(n+1- a)}{2} $$
The sum of natural squares\( : \sum k^2\)
The sum of the \( (n + 1) \) first natural squares, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^2 = \hspace{0.2em} 1 + \hspace{0.2em} 2^2 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (n-1)^2 + n^2 \hspace{0.1em} \hspace{0.1em} = \frac{n(n+1)(2n+1)}{6} $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k^2 = \frac{1}{6} \Bigl[ n+1-a\Bigr] \biggl[ n(2n+1) + a(2n +2a -1)\biggr] $$
The sum of natural cubes\( : \sum k^3\)
The sum of the \( (n + 1) \) first natural cubes, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^3 = \hspace{0.2em} 1 + \hspace{0.2em} 2^3 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (n-1)^3 + n^3 \hspace{0.1em} \hspace{0.1em} = \frac{n^2 (n+1)^2 }{4} $$
Moreover,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^3 = \Biggl( \hspace{0.1em} \sum_{k = 0}^n k \hspace{0.1em} \Biggr)^2$$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k^3 = \frac{1}{4} \Bigl[ n+1-a\Bigr] \biggl[ n^2(n+1) + a(n^2 + an +a^2 - a)\biggr] $$
The sum of odd numbers\( : \sum (2k +1) \)
An odd number \( O \) can be formulated as:
$$ \forall k \in \mathbb{Z}, \enspace O = 2k +1 $$
The sum of the \( (n + 1) \) first odd numbers, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n (2k +1) = 1 + 3 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (2n+1) = (n+1)^2 $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} (2k +1) = (n + 1+ a)(n+1- a) $$
The sum of even numbers\( : \sum (2k) \)
An even number \( E \) can be formulated as:
$$ \forall k \in \mathbb{Z}, \enspace E = 2k $$
The sum of the \( (n + 1) \) first even numbers, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n (2k) = 2 + 4 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} 2n = n(n+1) $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} (2k) = (n + a)(n+1- a) $$
The sum of the terms of an arithmetical sequence\( : \sum (u_0 + kr) \)
An arithmetical sequence is defined by:
$$ \forall n \in \mathbb{N}, \enspace \forall (u_0, r) \in \hspace{0.05em} \mathbb{R}^2, \enspace u_n = u_0 + nr $$
The sum of the \( (n + 1) \) first terms of an arithmetical sequence, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, \enspace \forall (u_0, r) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ \sum_{k = 0}^n (u_0 + kr) = \Bigl[n + 1 \Bigr]\Biggl[ \frac{u_0 + u_n}{2} \Biggr]$$
$$ (with \enspace u_n = u_0 + nr) $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k = a}^n (u_0 + kr) = \Bigl[n + 1 - a\Bigr] \Biggl[ \frac{u_0 + u_{n+a}}{2} \Biggr]$$
$$ (with \enspace u_n = u_0 + nr) $$
The sum of the natural powers of a real number\( : \sum q^k\)
The sum of the \( (n + 1) \) first natural powers of a real number, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, \enspace \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k = 0}^n q^k = 1 + q + q^2 \enspace + ... + \enspace q^{n-1} + q^n = \frac{q^{n+1} - 1}{q-1} $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, \enspace \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k =a}^{n} q^k = \frac{q^{n+1} - q^{a}}{q-1} $$
The sum of the terms of a geometrical sequence\( : \sum (v_0 . q^k) \)
A geometrical sequence is defined by:
$$ \forall n \in \mathbb{N}, \ \forall (v_0,\ q) \in \hspace{0.05em} \mathbb{R}^2 , \enspace v_n = v_0.q^n $$
The sum of the \( (n + 1) \) first terms of a geometrical sequence, that is to say from \(0\) to \(n\), is worth:
$$ \forall n \in \mathbb{N}, \enspace \forall v_0 \in \hspace{0.05em} \mathbb{R}, \ \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0.\frac{q^{n+1} - 1}{q-1} $$
In the general way, this sum going from \((k = a) \) until \(n\) is worth:
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, \enspace \forall v_0 \in \hspace{0.05em} \mathbb{R}, \ \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k =a}^{n} v_0.q^k = v_0.\frac{q^{n+1} - q^{a}}{q-1} $$
Let \((a,n) \in \hspace{0.05em} \mathbb{N}^2\) be two natural numbers.
We want to calculate the sum of the first natural numbers from \( 0\) to \( n\):
$$ \sum_{k = 0}^n k = \enspace \underbrace{0 + 1 + 2 \ + \ ... \ + \ (n-1) + n} _\text{(n+1) terms} $$
By definition, this sum is worth:
$$ \sum_{k = 0}^n k = \enspace \underbrace{0 + 1 + 2 \ + \ ... \ + \ (n-1) + n} _\text{(n+1) terms} $$
Let us rewrite this sum in reverse:
$$ \sum_{k = 0}^n k = n + (n-1) \enspace + \ ... \ + \enspace 2 + 1 + 0 $$
Additionning both equations by matching the terms one by one, we do have:
$$ \sum_{k = 0}^n k + \sum_{k = 0}^n k = \enspace \underbrace{(0 + n) + (1 + n - 1) \ + \ ... \ + \ (n - 1 + 1) + (n + 0)} _\text{(n+1) terms} $$
All these terms are worth \( n \):
$$ \sum_{k = 0}^n k + \sum_{k = 0}^n k = \enspace \underbrace{ n + n \ + \ ... \ + \ n + n} _\text{(n+1) terms} $$
Thus,
$$ 2\sum_{k = 0}^n k = n(n+1) $$
And as a result,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k = \frac{n(n+1)}{2} $$
We know from remarkable identities that:
$$ (k + 1)^2 = k^2 + 2k + 1$$
$$ (k + 1)^2 - k^2 - 1= 2k $$
Let us do the sum of each terms of both members of the equation, from \( (k =0) \) to \(n\).
$$ \sum_{k=0}^n \Bigl[ (k + 1)^2 - k^2 \Bigr] - \sum_{k=0}^n 1= \sum_{k=0}^n 2k $$
Now, we know that a telescoping of terms is going to happen, and it will remain only:
$$\sum_{k=0}^n \bigl [ a_{k+1} - a_k \bigr] = \underbrace{a_{n+1}} _\text{first term} - \underbrace{a_{0}} _\text{last term} $$
So in our case:
$$\sum_{k=0}^n \Bigl[(k+1)^2 - k^2 \Bigr] = (n + 1)^2 - \ 0^2 $$
Which leads us to:
$$ (n + 1)^2 -(n+1) = 2\sum_{k=0}^n k $$
$$ (n+1)(n+1 - 1) = 2\sum_{k=0}^n k $$
$$ \frac{n(n+1)}{2} = \sum_{k = 0}^n k$$
And as a result,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k = \frac{n(n+1)}{2} $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n k = \enspace \underbrace{a + (a + 1) + (a + 2) \ + \ ... \ + \ (n-1) + n} _\text{(n+1) terms} $$
$$ (k + 1)^2 = k^2 + 2k + 1$$
$$ (k + 1)^2 - k^2 - 1= 2k $$
$$ \sum_{k=a}^n \Bigl[ (k + 1)^2 - k^2 \Bigr] - \sum_{k=a}^n 1= \sum_{k=a}^n 2k $$
$$ (n + 1)^2 - a^2 - (n+1 -a )= 2\sum_{k=a}^n k $$
Factorizing the two first members of left part, we do have:
$$ (n + 1 - a)(n+1+a) - (n+1 -a )= 2\sum_{k=a}^n k $$
Factorizing now all the left part we do obtain:
$$ (n + 1 - a)(n+a)= 2\sum_{k=a}^n k $$
$$ \frac{(n + 1 - a)(n+a)}{2}= \sum_{k=a}^n k $$
Finally we do have,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k = \frac{(n + a)(n+1- a)}{2} $$
We want to calculate the sum of the first natural squares from \( 0\) to \( n\):
$$ \sum_{k = 0}^n k^2 = \enspace \underbrace{0 + 1 + 4 + 9 \ + \ ... \ + \ (n-1)^2 + n^2} _\text{(n+1) terms} $$
Thanks to the Newton's binomial, we know that:
$$\forall k \in \hspace{0.05em} \mathbb{R},$$
$$ (k + 1)^3 = \binom{3}{0} k^3 + \binom{3}{1}k^2 + \binom{3}{2}k + \binom{3}{3}$$
So,
$$ (k + 1)^3 = k^3 + 3k^2 + 3k + 1$$
$$ (k + 1)^3 - k^3 - 3k -1 = 3k^2 $$
Let us do the sum of each terms of both members of the equation, from \( (k =0) \) to \(n\).
Now, we know that a telescoping of terms is going to happen, and it will remain only:
$$\sum_{k=0}^n \bigl [ a_{k+1} - a_k \bigr] = a_{n+1} - a_{0} $$
So in our case:
$$\sum_{k=0}^n \Bigl[(k+1)^3 -k^3 \Bigr] = (n + 1)^3 - 0 $$
Which leads us to:
$$ (n + 1)^3 -(n+1) -3\sum_{k=0}^n k = 3\sum_{k=0}^n k^2 $$
The sum of the first natural numbers had been calculated above. Let us replace it.
$$ (n + 1)^3 -(n+1) -3 \Biggl[ \frac{n(n+1)}{2} \Biggr] = 3\sum_{k=0}^n k^2 $$
$$ (n+1)\Bigl((n + 1)^2 -1 - \frac{3n}{2} \Bigr) = 3\sum_{k=0}^n k^2 $$
$$ \frac{1}{2}(n+1)\Bigl(2n^2 +4n - 3n\Bigr) = 3\sum_{k=0}^n k^2 $$
$$ \frac{1}{6}(n+1)\Bigl(2n^2 +n\Bigr) = \sum_{k=0}^n k^2 $$
$$ \frac{n(n+1)(2n+1)}{6} = \sum_{k=0}^n k^2 $$
And as a result,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^2 = \frac{n(n+1)(2n+1)}{6} $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n k^2 = \enspace \underbrace{a^2 + (a + 1)^2 + (a + 2)^2 \ + \ ... \ + \ (n-1)^2 + n^2} _\text{(n+1) terms} $$
$$ (k + 1)^3 = k^3 + 3k^2 + 3k + 1$$
$$ (k + 1)^3 - k^3 -3 k - 1 = 3k^2 $$
$$ \sum_{k=a}^n \Bigl[(k + 1)^3 - k^3 \Bigr] -\sum_{k=a}^n 3 k -\sum_{k=a}^n 1 = \sum_{k=a}^n 3k^2 $$
Let us replace all sums by its value:
$$ (n+1)^3 - a^3 - 3\frac{(n + a)(n+1- a)}{2} -(n+1-a) = 3\sum_{k=a}^n k^2 $$
We know from the Bernouilli's formula (or geometrical identity) that:
$$ \forall (a,b) \in \hspace{0.05em} \mathbb{R}^2, \enspace \forall n \in \hspace{0.05em} \mathbb{N}, $$
$$a^n - b^n = (a-b) \sum_{k=0}^{n-1} a^{n-k-1}b^k $$
So in our case:
$$(n+1)^3 - a^3 = (n+1-a) \sum_{k=0}^{2} (n+1)^{2-k}a^k $$
$$(n+1)^3 - a^3 = (n+1-a) \Bigl[ (n+1)^2 + a(n+1) +a^2 \Bigr]$$
Which leads us to:
$$ (n+1-a) \Bigl[ (n+1)^2 + a(n+1) +a^2 \Bigr] - 3\frac{(n + a)(n+1- a)}{2} -(n+1-a) = 3\sum_{k=a}^n k^2 $$
We can now factorize it:
$$ (n+1-a) \Bigl[(n+1)^2 + a(n+1) +a^2 - 3\frac{(n + a)}{2} -1 \Bigr] = 3\sum_{k=a}^n k^2 $$
$$ \frac{ (n+1-a)}{2} \Bigl[2n^2 + 4n + 2 + 2an +2a + 2a^2 -3n -3a -2 \Bigr] = 3\sum_{k=a}^n k^2 $$
$$ \frac{ (n+1-a)}{2} \Bigl[2n^2 + n + a(2n+ 2a - 1 ) \Bigr] = 3\sum_{k=a}^n k^2 $$
And finally we obtain,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k^2 = \frac{1}{6} \Bigl[ n+1-a\Bigr] \biggl[ n(2n+1) + a(2n +2a -1)\biggr] $$
We want to calculate the sm of the first natural cubes from \( 0\) to \( n\):
$$ \sum_{k = 0}^n k^3 = \enspace \underbrace{0 + 1 + 8 + 27 \ + \ ... \ + \ (n-1)^3 + n^3} _\text{(n+1) terms} $$
We know from the Newton's binomial that:
$$\forall k \in \hspace{0.05em} \mathbb{R},$$
So,
$$ (k + 1)^4 = \binom{4}{0} k^4 + \binom{4}{1}k^3 + \binom{4}{2}k^2 + \binom{4}{3}k + 1$$
And,
$$ (k + 1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$
$$ (k + 1)^4 - k^4 - 6k^2 -4k -1 = 4k^3 $$
Let us do the sum of each terms of both members of the equation, from \( (k =0) \) to \(n\).
Now, we know that a telescoping of terms is going to happen, and it will remain only:
$$\sum_{k=0}^n \bigl [ a_{k+1} - a_k \bigr] = a_{n+1} - a_{0} $$
So in our case:
$$\sum_{k=0}^n \Bigl[(k+1)^4 -k^4 \Bigr] = (n + 1)^4 - 0 $$
Which leads us to:
$$ (n + 1)^4 -(n+1) -6\sum_{k=0}^n k^2 -4\sum_{k=0}^n k = 4\sum_{k=0}^n k^3 $$
The sum of the first natural numbers and also the sum of the first natural squares had already been calculated. Let us replace both of it by their respective value.
$$ (n + 1)^4 -(n+1) -6 \Biggl[\frac{n(n+1)(2n+1)}{6} \Biggr] -4 \Biggl[\frac{n(n+1)}{2} \Biggr] = 4\sum_{k=0}^n k^2 $$
$$ (n+1)\Bigl((n + 1)^3 -1 -n(2n+1) -2n \Bigr) = 3\sum_{k=0}^n k^2 $$
$$ (n+1)\Bigl(n^3 +3n^2 + 3n +1 -1 -2n^2 -n -2n^2 -2n \Bigr) = 3\sum_{k=0}^n k^2 $$
$$ (n+1)(n^3+ n^2) = 3\sum_{k=0}^n k^2 $$
And finally,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^3 = \frac{n^2 (n+1)^2 }{4} $$
Furthermore, we can notice that:
$$ \sum_{k = 0}^n k^3 = \frac{n^2 (n+1)^2 }{4} = \Biggl(\frac{n(n+1)}{2} \Biggr)^2 $$
So,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n k^3 = \Biggl( \hspace{0.1em} \sum_{k = 0}^n k \hspace{0.1em} \Biggr)^2$$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n k^3 = \enspace \underbrace{a^3 + (a + 1)^3 + (a + 2)^3 \ + \ ... \ + \ (n-1)^3 + n^3} _\text{(n+1) terms} $$
$$ (k + 1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$
$$ (k + 1)^4 - k^4 -6k^2 - 4k - 1 = 4k^3 $$
Let us do the sum of each terms of both members of the equation, from \( (k =0) \) to \(n\).
$$ \sum_{k=a}^n \Bigl[(k + 1)^3 - k^4 \Bigr] - \sum_{k=a}^n 6 k^2 - \sum_{k=a}^n 4k - \sum_{k=a}^n 1 = \sum_{k=a}^n 4k^3 $$
We already calculated both sums \(\sum k^2\) and \(\sum k\). Let us replace it by their respective value.
$$ (n+1)^4 - a^4 - 6 \times \frac{1}{6} \Bigl[ n+1-a\Bigr] \biggl[ n(2n+1) + a(2n +2a -1)\biggr] -4 \frac{(n + a)(n+1- a)}{2} -(n+1-a) = 4\sum_{k=a}^n k^3 $$
We use again the Bernouilli's formula, in our case it will be:
$$(n+1)^4 - a^4 = (n+1-a) \sum_{k=0}^{3} (n+1)^{3-k}a^k $$
$$(n+1-a) \Bigl[ (n+1)^3 + a^2(n+1) +a(n+1)^2 + a^3 \Bigr]$$
Which leads us to:
$$(n+1-a) \Bigl[ (n+1)^3 + a^2(n+1) +a(n+1)^2 + a^3 \Bigr] - \Bigl[ n+1-a\Bigr] \biggl[ n(2n+1) + a(2n +2a -1)\biggr] -2 (n + a)(n+1- a) -(n+1-a) = 4\sum_{k=a}^n k^3 $$
We can know factorize all:
$$(n+1-a) \Biggl[ (n+1)^3 + a^2(n+1) +a(n+1)^2 + a^3 - \biggl[ n(2n+1) + a(2n +2a -1)\biggr] -2(n + a) -1 \Biggr] = 4\sum_{k=a}^n k^3 $$
$$(n+1-a) \Bigl[ n^3 + 3n^2 + 3n + 1 + a^2n + a^2 +an^2 + 2an+a+a^3 - 2n^2 -n -2an -2a^2 +a -2n -2a \Bigr] = 4\sum_{k=a}^n k^3 $$
$$(n+1-a) \biggl[n^2(n+1) + a(n^2 + an +a^2 - a)\biggr] = 4\sum_{k=a}^n k^3 $$
And finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} k^3 = \frac{1}{4} \Bigl[ n+1-a\Bigr] \biggl[ n^2(n+1) + a(n^2 + an +a^2 - a)\biggr] $$
An odd number \( O \) can be formulated as:
$$ \forall k \in \mathbb{Z}, \enspace O = 2k +1 $$
We want to calculate the first odd number from \( 0\) to \( n\):
$$ \sum_{k = 0}^n (2k +1) = \enspace \underbrace{1 + 3 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (2n+1) } _\text{(n+1) terms} $$
We can split this sum into two differents:
$$ \sum_{k = 0}^n (2k +1) = \sum_{k = 0}^n (2k) + \sum_{k = 0}^n 1 $$
$$ \sum_{k = 0}^n (2k +1) = 2\sum_{k = 0}^n k + \sum_{k = 0}^n 1 $$
The sum of the first natural numbers had been calculated above, let inject it:
$$ \sum_{k = 0}^n (2k +1) = n(n+1) + (n+1) $$
$$ \sum_{k = 0}^n (2k +1) = (n+1)(n+1) $$
And as a result,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n (2k +1) = (n+1)^2 $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n (2k +1) = \enspace \underbrace{ (2a+1) + \Bigl[2(a+1) +1 \Bigr] \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (2n+1) } _\text{(n+1 -a) terms} $$
$$ \sum_{k = a}^n (2k +1) = \sum_{k = a}^n (2k) + \sum_{k = a}^n 1 $$
$$ \sum_{k = a}^n (2k +1) = 2\sum_{k = a}^n k + \sum_{k = a}^n 1 $$
$$ \sum_{k = a}^n (2k +1) = (n + a)(n+1- a) + (n+1- a) $$
$$ \sum_{k = a}^n (2k +1) = (n+1- a) (n + 1 +a) $$
And finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} (2k +1) = (n + 1 + a)(n+1- a) $$
An even number \( E \) can be formulated as:
$$ \forall k \in \mathbb{Z}, \enspace E = 2k $$
We want to calculate the sum of the first even numbers from \( 0\) to \( n\).
$$ \sum_{k = 0}^n 2k = \enspace \underbrace{2 +4 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} 2n } _\text{(n+1) terms} $$
$$ \sum_{k = 0}^n 2k = 2\sum_{k = 0}^n k $$
The sum of the first natural numbers had been calculated above, let inject it:
$$ \sum_{k = 0}^n 2k = n(n+1) $$
And finally,
$$ \forall n \in \mathbb{N}, $$
$$ \sum_{k = 0}^n 2k= n(n+1) $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n 2k = \enspace \underbrace{2a +4a \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} 2n } _\text{(n+1) terms} $$
$$ \sum_{k = a}^n 2k = 2\sum_{k = a}^n k = (n + a)(n+1- a) $$
And finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^{n} (2k) = (n + a)(n+1- a) $$
Let \( (u_n)_{n \in \mathbb{N}}\) be an arithmetical sequence with a reason \( r \in \mathbb{R}\) and its first term \( u_0\in \mathbb{R}\).
We want to calculate the sum of the terms of this sequence from \( 0\) to \( n\):
$$ \sum_{k = 0}^n (u_0 + kr) = u_0 + (u_0 + r) + (u_0 + 2r) \enspace + \enspace ... \enspace + \enspace \Bigl[u_0 + (n -1)r \Bigr] + (u_0 + nr) $$
So,
$$ \sum_{k = 0}^n (u_0 + kr) = u_0(n + 1) + r \bigl(1 + 2 \enspace + \enspace ... \enspace + \enspace (n -1) + n \bigr) $$
$$ \sum_{k = 0}^n (u_0 + kr) = u_0(n + 1) + r \Biggl[ \sum_{k = 0}^n k \Biggr] $$
The sum of the first natural numbers had been calculated above, let inject it:
$$ \sum_{k = 0}^n (u_0 + kr) = u_0(n + 1) + r\Biggl[\frac{n(n + 1)}{2}\Biggr]$$
Now, factorizing by \( (n + 1) \) we obtain:
$$ \sum_{k = 0}^n (u_0 + kr) = \Bigl[n + 1 \Bigr]\Biggl[u_0 + \frac{ nr}{2}\Biggr] $$
$$ \sum_{k = 0}^n (u_0 + kr) = \Bigl[n + 1 \Bigr]\Biggl[\frac{ 2 u_0 + nr}{2}\Biggr] $$
$$ \sum_{k = 0}^n (u_0 + kr) = \Bigl[n + 1 \Bigr]\Biggl[\frac{ u_0 + u_n}{2}\Biggr] $$
And finally,
$$ \forall n \in \mathbb{N}, \enspace \forall (u_0, r) \in \hspace{0.05em} \mathbb{R}^2, $$
$$ \sum_{k = 0}^n (u_0 + kr) = \Bigl[n + 1 \Bigr]\Biggl[ \frac{u_0 + u_n}{2} \Biggr]$$
$$ (with \enspace u_n = u_0 + nr) $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n (u_0 + kr) = (u_0 + ar) + \Bigl[u_0 + (a+1)r \Bigl] \enspace + \enspace... \enspace + \enspace \Bigl[u_0 + (n -1)r \Bigl] + (u_0 + nr) $$
So,
$$ \sum_{k = a}^n (u_0 + kr) = u_0(n + 1-a) + r \Bigl[a + (a+1) \enspace + \enspace ... \enspace + \enspace (n -1) + n \Bigr] $$
$$ \sum_{k = a}^n (u_0 + kr) = u_0(n + 1-a) + r \Biggl[ \sum_{k = a}^n k \Biggr] $$
$$ \sum_{k = a}^n (u_0 + kr) = u_0(n + 1-a) + r \Biggl[\frac{ (n + a)(n+1- a)}{2}\Biggr] $$
$$ \sum_{k = a}^n (u_0 + kr) = \Bigl[n + 1 - a\Bigr] \Biggl[ u_0 + \frac{ (n+a)r}{2}\Biggr] $$
$$ \sum_{k = a}^n (u_0 + kr) = \Bigl[n + 1 - a\Bigr] \Biggl[ \frac{ 2u_0 + (n+a)r}{2}\Biggr] $$
$$ \sum_{k = a}^n (u_0 + kr) = \Bigl[n + 1 - a\Bigr] \Biggl[ \frac{ u_0 + u_{n+a}}{2}\Biggr] $$
So finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, $$
$$ \sum_{k =a}^n (u_0 + kr) = \Bigl[n + 1 - a\Bigr] \Biggl[ \frac{u_0 + u_{n+a}}{2} \Biggr]$$
$$ (with \enspace u_n = u_0 + nr) $$
Let \( q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , \) be a real number.
We want to calculate the sum of the first natural powers of a real \(q\) from \( 0\) to \( n\):
$$ \sum_{k = 0}^n q^k = \enspace \underbrace{1 + q + q^2 \enspace + ... + \enspace q^{n-1} + q^n} _\text{(n+1) terms} \qquad (1) $$
Multuplying all by \( q \) we obtain:
$$ q.\left[\sum_{k = 0}^n q^k \right] = q.(1 + q + q^2 \enspace + ... + \enspace q^{n-1} + q^n) $$
We now develop the right member:
$$ q.\left[\sum_{k = 0}^n q^k \right] = q + q^2 + q^3 \enspace + ... + \enspace q^n + q^{n+1} \qquad (2) $$
By substracting \( (1) \) from \( (2) \), we realize that a telescoping of terms occurs:
$$ q.\left[\sum_{k = 0}^n q^k \right] - \left[\sum_{k = 0}^{n} q^k \right] = q^{n+1} - 1 $$
Finally, we factorize by \( \sum q^k \) and:
$$ (q - 1).\left[\sum_{k = 0}^n q^k \right] = q^{n+1} - 1 $$
As a result we do have,
$$ \forall n \in \mathbb{N}, \enspace \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k = 0}^n q^k = \frac{q^{n+1} - 1}{q-1} $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n q^k = \enspace \underbrace{q^a + q^{a+1} + q^{a+2} \enspace + ... + \enspace q^{n-1} + q^n} _\text{(n+1- a) terms} $$
$$ q \Biggl[ \sum_{k = a}^n q^k \Biggr] = q(q^a + q^{a+1} + q^{a+2} \enspace + ... + \enspace q^{n-1} + q^n) $$
$$ q \Biggl[ \sum_{k = a}^n q^k \Biggr] = q^{a+1} + q^{a+2} + q^{a+3} \enspace + ... + \enspace q^{n} + q^{n+1} $$
$$ q.\left[\sum_{k = a}^n q^k \right] - \left[\sum_{k = a}^{n} q^k \right] = q^{n+1} - q^a $$
$$ \sum_{k = a}^n q^k = \frac{q^{n+1} - q^a }{q-1} $$
And finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, \enspace \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k =a}^{n} q^k = \frac{q^{n+1} - q^{a}}{q-1} $$
Let \( (v_n)_{n \in \mathbb{N}}\) be a geometrical sequence with a reason \( q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] \) and its first term \( v_0\in \mathbb{R}\).
We want to calculate the sum of the terms of this sequence from \( 0\) to \( n\):
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0 + v_0.q + v_0.q^2 \enspace + ... + \enspace + v_0.q^{n - 1} + v_0.q^n $$
So,
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0.\Biggl[ \sum_{k = 0}^n q^k \Biggr] $$
The sum of the first natural powers of a real number has already been calculated, let us inject it:
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0.\frac{q^{n+1} - 1}{q-1} $$
And finally,
$$ \forall n \in \mathbb{N}, \enspace \forall v_0 \in \hspace{0.05em} \mathbb{R}, \ \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0.\frac{q^{n+1} - 1}{q-1} $$
Let us do the same reasoning as before, but now from \((k = a) \) until \(n\):
$$ \sum_{k = a}^n (v_0.q^k ) = v_0q^a + v_0.q^{a+1} + v_0.q^{a+2} \enspace + ... + \enspace + v_0.q^{n - 1} + v_0.q^n $$
$$ \sum_{k = a}^n (v_0.q^k ) = v_0.\Biggl[ \sum_{k = a}^n q^k \Biggr] $$
$$ \sum_{k = 0}^n (v_0.q^k ) = v_0.\frac{q^{n+1} - q^{a}}{q-1}$$
And finally,
$$ \forall (a,n) \in \hspace{0.05em} \mathbb{N}^2, \enspace \forall v_0 \in \hspace{0.05em} \mathbb{R}, \ \forall q \in \hspace{0.05em} \Bigl[ \mathbb{R} \ \backslash \bigl \{0,1 \bigr \} \Bigr] , $$
$$ \sum_{k =a}^{n} v_0.q^k = v_0.\frac{q^{n+1} - q^{a}}{q-1} $$
Let us calculate the following sum \(S\):
$$ S= \sum_{k =5}^{20} k = 5 + 6 + 7 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} 20 $$
$$ \sum_{k =0}^{n} k = \frac{n(n+1)}{2} $$
So,
$$ S = \Biggl[\sum_{k =0}^{20} k\Biggr] - \Biggl[\sum_{k =0}^{4} k \Biggr]$$
$$ S = \frac{20 \times 21 }{2} - \frac{4 \times 5 }{2} $$
$$ S = 210- 10 = 200 $$
At this stage, we will use the general formula:
$$ \sum_{k =a}^{n} k = \frac{(n + a)(n+1- a)}{2} $$
So,
$$ S = \sum_{k =5}^{20} k = \frac{(20 + 5)(20+1- 5)}{2} $$
$$ S = \sum_{k =5}^{20} k = \frac{25 \times 16}{2} = 200$$
Let us calculate the following sum \(S'\):
$$ S'= \sum_{k =7}^{11} k^3 = \ 7^{3} + \ 8^{3} + \ 9^3 + \ 10^3 + \ 11^3 $$
$$ \sum_{k = 0}^n k^3 = \frac{n^2 (n+1)^2 }{4} $$
So,
$$ S' = \Biggl[\sum_{k = 0}^{11} k^3 \Biggr] - \Biggl[\sum_{k = 0}^6 k^3 \Biggr]$$
$$ S' = \frac{11^2 (11+1)^2 }{4}- \frac{6^2 (6+1)^2 }{4}$$
$$ S' = \frac{121 \times 144}{4}- \frac{ 36 \times 49 }{4} = 3915$$
$$ \sum_{k =a}^{n} k^3 = \frac{1}{4} \Bigl[ n+1-a\Bigr] \biggl[ n^2(n+1) + a(n^2 + an +a^2 - a)\biggr] $$
So,
$$ S' = \sum_{k =7}^{11} k^3 = \frac{1}{4} \Bigl[ 11+1- 7\Bigr] \biggl[ 11^2 \times (11+1) + 7 \times (11^2 + \ 7 \times 11 + \ 7^2 - \ 7)\biggr] $$
$$ S' = \sum_{k =7}^{11} k^3 = \frac{5}{4} \times \biggl[ 121 \times 12 + 7 \times ( 121 + 77 + 49 - 7)\biggr] $$
$$ S' = \sum_{k =7}^{11} k^3 = \frac{5}{4} \times \biggl[ 1452 + 1680 \biggr] = 3915 $$
Go to the top of the page
