133
$\begingroup$

I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

$\endgroup$
8

32 Answers 32

1
2
-1
$\begingroup$

If you know that the nth square is the sum of the first n odd numbers, you can rewrite each square in the above sum in that way and do a little bit of rearranging to get the desired identity. In general, knowing the value of (n + 1)^k - n^k allows you to write (n+1)^k as a telescoping sum of a polynomial of degree k-1, running over the values 1 through n. If you know the values of the sums of consecutive powers up to k-1, this allows you to find the sum of consecutive kth powers by substituting the polynomial sum for each kth power.

$\endgroup$
-2
$\begingroup$

Another method by using telescoping sum :- We know $(a+b)^3-a^3-b^3=3ab(a+b)$ , take

$a=k-1 , b=2$ , then $a+b=k+1$ and $(k+1)^3-(k-1)^3-2^3=6(k-1)(k+1)=6k^2-6$ ,

hence $(k+1)^3-(k-1)^3-8+6=(k+1)^3-k^3+k^3-(k-1)^3-2=6k^2$ , taking sum over $k$

from $1$ to $n$ we get , $\sum_{k=1}^n [(k+1)^3-k^3] + \sum_{k=1}^n [k^3-(k-1)^3] -\sum_{k=1}^n 2 = 6 \sum_{k=1}^nk^2$ , the first

sum on the left hand side is telescoping resulting in $(n+1)^3-1$ , the second sum is also telescoping resulting in $n^3-(1-1)^3=n^3$ , and the third sum is simply $2n$ , hence

$6 \sum_{k=1}^nk^2=(n+1)^3-1+n^3-2n=2n^3+3n^2+n=n(n+1)(2n+1)$

$\endgroup$
1
2

Not the answer you're looking for? Browse other questions tagged or ask your own question.