sigma n =n²×a^-n.solve
Answers
The series \sum\limits_{k=1}^n k^a = 1^a + 2^a + 3^a + \cdots + n^a
k=1
∑
n
k
a
=1
a
+2
a
+3
a
+⋯+n
a
gives the sum of the a^\text{th}a
th
powers of the first nn positive numbers, where aa and nn are positive integers. Each of these series can be calculated through a closed-form formula. The case a=1,n=100a=1,n=100 is famously said to have been solved by Gauss as a young schoolboy: given the tedious task of adding the first 100100 positive integers, Gauss quickly used a formula to calculate the sum of 5050.5050.
The formulas for the first few values of aa are as follows:
\begin{aligned} \sum_{k=1}^n k &= \frac{n(n+1)}2 \\ \sum_{k=1}^n k^2 &= \frac{n(n+1)(2n+1)}6 \\ \sum_{k=1}^n k^3 &= \frac{n^2(n+1)^2}4. \end{aligned}
k=1
∑
n
k
k=1
∑
n
k
2
k=1
∑
n
k
3
=
2
n(n+1)
=
6
n(n+1)(2n+1)
=
4
n
2
(n+1)
2
.
Faulhaber's formula, which is derived below, provides a generalized formula to compute these sums for any value of a.a.