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Sum to n terms of 12 + 32 + 52 + …………….. isA) \(\cfrac{n(n^2-4)}6\)B) \(\cfrac{n(4n-1)}6\)C) \(\cfrac{n(n^2-4)}3\)D) \(\cfrac{n(4n^2-1)}3\) |
Answer» Correct option is (D) \(\frac{n(4n^2-1)}{3}\) \(S_n=1^2+3^2+5^2+......+\) upto n terms \(=1^2+3^2+5^2+......+(2n-1)^2\) \(=\sum(2n-1)^2\) \(=\sum(4n^2-4n+1)\) \(=4\sum n^2-4\sum n+\sum1\) \(=\frac{4n(n+1)(2n+1)}6-\frac{4n(n+1)}2+n\) \(=n\left(\frac23(2n^2+3n+1)-2(n+1)+1\right)\) \(=\frac n3(4n^2+6n+2-6n-6+3)\) \(=\frac n3(4n^2-1)\) Correct option is D) \(\cfrac{n(4n^2-1)}3\) |
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