How to simplify $1^2+2^2+3^2+\cdots+n^2=？$

Question

How to simplify $1^2+2^2+3^2+\cdots+n^2= \text{?}$ help…………………… thank you

Answer 1

$$\sum n^2= \frac{(n)(n+1)(2n+1)}{6}$$ thats the simplification nothing much in it.

Answer 2

It's equals to $$\frac{n(n+1)(2n+1)}{6}$$ It can be proven by using induction mehod.

Answer 3

We can evaluate the sum a number of ways. Here, we present two approaches, both of which rely on the sum of the arithmetic progression $\sum_{k=1}^nk=\frac12n(n+1)$.

METHODOLOGY $1$:

Here, we note that $\sum_{j=1}^k(1)=k$. Then, we can write

$$\begin{align} \sum_{k=1}^nk^2&=\sum_{k=1}^nk\sum_{j=1}^k(1)\\\\ &=\sum_{j=1}^n\sum_{k=j}^n\,k\\\\ &=\frac12\sum_{j=1}^n(n+1-j)(j+n)\\\\ &=\frac12\sum_{j=1}^n\left(n(n+1)+j-j^2\right)\\\\ &=\frac12n^2(n+1)+\frac14n(n+1)-\frac12\sum_{j=1}^nj^2\\\\ \frac32\sum_{k=1}^nk^2&=\frac{(2n+1)n(n+1)}{4}\\\\ \sum_{k=1}^nk^2&=\frac{(2n+1)n(n+1)}{6} \end{align}$$

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METHODOLOGY $2$: Making use of a telescoping sum

Note that we have $\sum_{k=1}^n\left((k+1)^3-k^3\right)=(n+1)^3-1$ since the summands are telescoping. Then, by expanding the term $(n+1)^3$, we can write

$$\begin{align} (n+1)^3-1&=\sum_{k=1}^n\left((k+1)^3-k^3\right)\\\\ &=\sum_{k=1}^n\left(3k^2+3k+1\right)\\\\ &=3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n(1)\\\\ &=3\sum_{k=1}^nk^2+\frac{3n(n+1)}{2}+n\\\\ \frac{(n+1)^3-1-\frac{3n(n+1)}{2}-n}{3}&=\sum_{k=1}^nk^2\\\\ \frac{(2n+1)n(n+1)}{6}&=\sum_{k=1}^nk^2 \end{align}$$

as expected!