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Question
Find 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms.
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Solution
2, 4, 6, … are in A.P.
∴ rth term = 2 + (r – 1) 2 = 2r
6, 9, 12, … are in A.P.
∴ rth term = 6 + (r – 1) (3) = (3r + 3)
∴ 2 x + 6 + 4 x 9 + 6 x 12 + ... upto n terms
= \[\displaystyle\sum_{r=1}^{n} 2r \times (3r + 3)\]
= 6\[\displaystyle\sum_{r=1}^{n}r^2 + 6\displaystyle\sum_{r=1}^{n}r\]
= `6*("n"("n" + 1)(2"n" + 1))/6 + 6("n"("n" + 1))/2`
= n(n + 1) (2n + 1 + 3)
= 2n(n + 1)(n + 2).
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