Sum to ‘n’ terms of 1, 8, 27, 64, ……………A) \(\cfrac{n^2(

1.	Sum to ‘n’ terms of 1, 8, 27, 64, ……………A) \(\cfrac{n^2(n+1)}4\)B) \(\cfrac{n^2(n+1)^2}4\)C) \(\cfrac{n(n+1)}2\)D) \(\cfrac{n(n+1)(2n+1)}6\)
Answer» Correct option is (B) \(\frac{n^2(n+1)^2}{4}\) \(1+8+27+64+......+n\) terms \(=1^3+2^3+3^3+4^3+.........+n\) terms \(=\left(\frac{n(n+1)}{2}\right)^2\) \(=\frac{n^2(n+1)^2}{4}\) Correct option is B) \(\cfrac{n^2(n+1)^2}4\)

Sum to ‘n’ terms of 1, 8, 27, 64, ……………A) \(\cfrac{n^2(n+1)}4\)B) \(\cfrac{n^2(n+1)^2}4\)C) \(\cfrac{n(n+1)}2\)D) \(\cfrac{n(n+1)(2n+1)}6\)

Answer»

Correct option is (B) \(\frac{n^2(n+1)^2}{4}\)

\(1+8+27+64+......+n\) terms

\(=1^3+2^3+3^3+4^3+.........+n\) terms

\(=\left(\frac{n(n+1)}{2}\right)^2\)

\(=\frac{n^2(n+1)^2}{4}\)

Correct option is B) \(\cfrac{n^2(n+1)^2}4\)