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0 votes
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in Arithmetic Progression by (43.0k points)
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Sum to n terms of 12 + 32 + 52 + …………….. is

A) \(\cfrac{n(n^2-4)}6\)

B) \(\cfrac{n(4n-1)}6\)

C) \(\cfrac{n(n^2-4)}3\)

D) \(\cfrac{n(4n^2-1)}3\)

2 Answers

+1 vote
by (57.0k points)
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Best answer

Correct option is (D) \(\frac{n(4n^2-1)}{3}\)

\(S_n=1^2+3^2+5^2+......+\) upto n terms

\(=1^2+3^2+5^2+......+(2n-1)^2\)

\(=\sum(2n-1)^2\)

\(=\sum(4n^2-4n+1)\)

\(=4\sum n^2-4\sum n+\sum1\)

\(=\frac{4n(n+1)(2n+1)}6-\frac{4n(n+1)}2+n\)

\(=n\left(\frac23(2n^2+3n+1)-2(n+1)+1\right)\)

\(=\frac n3(4n^2+6n+2-6n-6+3)\)

\(=\frac n3(4n^2-1)\)

+1 vote
by (43.3k points)

Correct option is D) \(\cfrac{n(4n^2-1)}3\)

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