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Evaluate the following :\(\displaystyle\sum_{n =2}^{10}\) 4n |
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Answer» \(\displaystyle\sum_{n =2}^{10}\) 4n Now this term is in GP. 16, 64, 256…to 10 terms ∴ Common Ratio = r = \(\frac{64}{16} = 4\) ∴ Sum of GP for n terms = \(\frac{a(r^n -1)}{r-1} \) ........(1) ⇒ a = 16, r = 4, n = 10 ∴ Substituting the above values in (1) we get ⇒ \(\cfrac{16(4^{10} - 1)}{4-1}\) ⇒ 5592400 |
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