Question

What is sum of the square of the first in Natural nubers Tequation with derivation

JEE/Engineering Exams

Maths

Solution

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(et ( S=sum_{r=1}^{n} r^{2} ) ( n^{3}-(n-1)^{3}=3 n^{2}-3 n+1 ) Substituting ( n=1,2,3,4,5 ) [ begin{array}{c} 1^{3}-0^{3}=3 cdot 1^{2}-3 cdot 1+1 2^{3}-1^{3}=3 cdot 2^{2}-3 cdot 2+1 n^{3}-(n-1)^{3}=3 cdot n^{2}-3 cdot n+1 end{array} ] Aading we get [ n^{3}-0^{3}=3left(1^{2}+2^{2}+3^{2}+4^{2}+cdots+n^{2}right)-3(1+2+3+n) ] [ begin{array}{l} quad+(1+1+1+1 cdots+n) n^{3}=35-3 frac{n(n+1)}{2}+n 3 S=frac{n(n+1)(2 n+1)}{2} S=frac{n(n+1)(2 n+1)}{6} end{array} ]