Evaluate the following : Σ_{n = 2}^10 4^n

Question

Evaluate the following :

\(\displaystyle\sum_{n =2}^{10}\) 4ⁿ

Answer 1

 \(\displaystyle\sum_{n =2}^{10}\) 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