Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+………………..(a) 784(b) 882(c) 928(d) 966I had been asked this question in a job interview.Asked question is from Sum to n Terms of Special Series in chapter Sequences and Series of Mathematics – Class 11

Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+……

1.	Find the sum to 6 terms of each of the series 23+46+611+818+………………..(a) 784(b) 882(c) 928(d) 966I had been asked this question in a job interview.Asked question is from Sum to n Terms of Special Series in chapter Sequences and Series of Mathematics – Class 11
Answer» Correct OPTION is (d) 966 Explanation: GENERAL term of above series is ak = 2K(k^2+2) = 2k^3+4k Taking summation from k=1 to k=n on both sides, we get \(\sum_{i=0}^na_k = 2\sum_{i=0}^nk^3 + 4\sum_{i=0}^nk = 2(\FRAC{n(n+1)}{2})^2 + 4\frac{n(n+1)}{2}\) = n^2(n+1)^2/2+2n(n+1) = 3649/2 + 267 = 966.

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Find the sum to 6 terms of each of the series 2*3+4*6+6*11+8*18+………………..(a) 784(b) 882(c) 928(d) 966I had been asked this question in a job interview.Asked question is from Sum to n Terms of Special Series in chapter Sequences and Series of Mathematics – Class 11

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Find the sum to 6 terms of each of the series 23+46+611+818+………………..(a) 784(b) 882(c) 928(d) 966I had been asked this question in a job interview.Asked question is from Sum to n Terms of Special Series in chapter Sequences and Series of Mathematics – Class 11