InterviewSolution
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InterviewSolution
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Sum to ‘n’ terms of 3.5 + 4.7 + 5.9 + ………………A) \(\cfrac{n(n+1)(n-3)}{12}\)B) \(\cfrac{n(4n^2+27n+59)}{6}\)C) \(\cfrac{n(25n^2+65n+2)}{12}\)D) \(\cfrac{n(25n^2+65n+2)}{12}\) |
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Answer» Correct option is (B) \(\frac{n(4n^2+27n+59)}{6}\) 3.5 + 4.7 + 5.9 + ……. + n terms = 3.5 + 4.7 + 5.9 + ……. + (n+2) (2n+3) \((\because\) 3, 4, 5, ........ is an arithmetic progression whose \(n^{th}\) term is \(a_n=n+2\) & 5, 7, 9, ...... is an A.P. whose \(n^{th}\) term is \(a_n=(2n+3))\) \(=\sum(n+2)(2n+3)\) \(=\sum(2n^2+4n+3n+6)\) \(=\sum(2n^2+7n+6)\) \(=2\sum n^2+7\sum n+6\sum 1\) \(=2\,\left(\frac{n(n+1)(2n+1)}6\right)+7\,\left(\frac{n(n+1)}2\right)+6n\) \(\left(\because\sum n^2=\frac{n(n+1)(2n+1)}6,\sum n=\frac{n(n+1)}2\;\&\;\sum1=n\right)\) \(=\frac n6(2(n+1)(2n+1)+21(n+1)+36)\) \(=\frac n6(4n^2+6n+2+21n+21+36)\) \(=\frac n6(4n^2+27n+59)\) Correct option is B) \(\cfrac{n(4n^2+27n+59)}{6}\) |
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