Question

*Find the formula of the series 1+2²+3²+4²+....+n²*​

Answer 1

The formula is $\frac{n(n+1)(2n+1)}{6}$

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let $S=1+2^2+3^2+4^2+................+n^2$

We know $(n+1)^3=n^3+3n^2+3n+1$

$(1+1)^3=1^3+3(1)^2+3(1)+1$

$(2+1)^3=2^3+3(2)^2+3(2)+1$

$(3+1)^3=3^3+3(3)^2+3(3)+1$

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$(n+1)^3=n^3+3(n)^2+3(n)+1$

On adding

$2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)$

$2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)$

$(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n$

$n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n$

$2n^3+6n^2+6n=6S+3n^2+3n+2n$

$6S=2n^3+3n^2+n$

$6S=2n^2(n+1)+n(n+1)$

$6S=(n+1)(2n^2+n)$

$6S=n(n+1)(2n+1)$

$S=\frac{n(n+1)(2n+1)}{6}$

Hence, the formula is $\frac{n(n+1)(2n+1)}{6}$

Learn more about Series here:

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