The formula is [tex]\frac{n(n+1)(2n+1)}{6}[/tex]
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 [tex]S=1+2^2+3^2+4^2+................+n^2[/tex]
We know [tex](n+1)^3=n^3+3n^2+3n+1[/tex]
[tex](1+1)^3=1^3+3(1)^2+3(1)+1[/tex]
[tex](2+1)^3=2^3+3(2)^2+3(2)+1[/tex]
[tex](3+1)^3=3^3+3(3)^2+3(3)+1[/tex]
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[tex](n+1)^3=n^3+3(n)^2+3(n)+1[/tex]
On adding
[tex]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)[/tex]
[tex]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)[/tex]
[tex](n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n[/tex]
[tex]n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n[/tex]
[tex]2n^3+6n^2+6n=6S+3n^2+3n+2n[/tex]
[tex]6S=2n^3+3n^2+n[/tex]
[tex]6S=2n^2(n+1)+n(n+1)[/tex]
[tex]6S=(n+1)(2n^2+n)[/tex]
[tex]6S=n(n+1)(2n+1)[/tex]
[tex]S=\frac{n(n+1)(2n+1)}{6}[/tex]
Hence, the formula is [tex]\frac{n(n+1)(2n+1)}{6}[/tex]
Learn more about Series here:
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