Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$

Question

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?

Answer 1

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 $k$th powers by substituting the polynomial sum for each $k$th power.

Answer 2

There are many ways to evaluate this sum, but I will use the approach that I came up with when I was in high school. This will be a natural approach that anyone can come up with.

$$S_n:=\sum_{k=1}^n k^2=\sum_{k=1}^n k\times k=\sum_{k=1}^nk(k-1)+\sum_{k=1}^nk$$ It is well known that $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$, so the real problem is $\sum\limits_{k=1}^nk(k-1)$. Since $\dbinom{n}{2} =\frac{n(n-1)}{2}$, then

$$\sum_{k=1}^nk(k-1)= 2\sum_{k=1}^n\frac{k(k-1)}{2}=2\sum_{k=3}^n \dbinom{k}{2}+\frac{1\times 0}{2}+ 1$$

Since $\displaystyle\dbinom{n}{2}+\dbinom{n}{3}= \frac{n!}{2!(n-3)!}\bigg(\frac{1}{n-2}+\frac{1}{3}\bigg)=\frac{n!}{2!(n-3)!}\bigg(\frac{n+1}{(n-2)(3)}\bigg)=\dbinom{n+1}{3} $, then

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

Now the term $\dbinom{3}{3}$ will "eat" $\sum\limits_{k=3}^n \dbinom{k}{2}$ as follows:

$\\[10pt]$ $$\dbinom{3}{3}+\sum_{k=3}^3\dbinom{k}{2}=\dbinom{3}{3}+\dbinom{3}{2}= \dbinom{4}{3}$$ $$\dbinom{3}{3}+\sum_{k=3}^4\dbinom{k}{2}=\dbinom{4}{3}+ \dbinom{4}{2}=\dbinom{5}{3}$$ $$\dbinom{3}{3}+\sum_{k=3}^5\dbinom{k}{2}=\dbinom{5}{3}+ \dbinom{5}{2}=\dbinom{6}{3}$$ $$.$$

$$.$$

$$.$$ $$\dbinom{3}{3}+\sum_{k=3}^n\dbinom{k}{2}=\dbinom{n}{3}+ \dbinom{n}{2}=\dbinom{n+1}{3}$$

$\\[10pt]$

So $\displaystyle\sum_{k=1}^n{k(k-1)}= 2\dbinom{n+1}{3}=\frac{(n+1)(n)(n-1)}{3}$

$$S_n= \frac{n(n+1)(n-1)}{3}+ \frac{n(n+1)}{2}= \frac{n(n+1)}{6 }\bigg( 2(n-1)+3 \bigg)= \frac{n(2n+1)(n+1)}{6}$$

---

It is worth mentioning that this approach can work for any $i \in \mathbb{N} $ to evaluate $\sum\limits_{k=1}^n k^i$. When I was in high school I was able to to find the sum up to $i=6$ by this approach.

Answer 3

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

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