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Sum of first n terms of 1.4 + 4.7 + 7.10 + 10.13 + ……………… is A) Σn2 – 4Σn + nB) 9Σn2 – 3Σn – 2n C) Σn2 – 3Σn + 4 D) 9Σn2 – 3Σn – 2 |
Answer» Correct option is (B) \(9 \sum n^2 - 3 \sum n - 2n\) 1, 4, 7, 10, ........ will form an A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\) \(=1+(n-1)3\) \(=3n-3+1\) \(=3n-2\) Also 4, 7, 10, 13, ......... will form another A.P. whose \(n^{th}\) term is \(a_n=a+(n-1)d\) \(=4+(n-1)3\) \(=3n-3+n\) \(=3n+1\) Now, 1.4+4.7+7.10+10.13+.........+n terms = 1.4+4.7+7.10+10.13+.........+(3n-2)(3n+1) \(=\sum(3n-2)(3n+1)\) \(=\sum(9n^2-6n+3n-2)\) \(=\sum(9n^2-3n-2)\) \(=9\sum n^2-3\sum n-2\sum1\) \(=9\sum n^2-3\sum n-2n\) \((\sum1=n)\) Correct option is B) 9Σn2 – 3Σn – 2n |
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