1.

Evaluate the following :\(\displaystyle\sum_{n =2}^{10}\) 4n

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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