Evaluate the following: `sum_(n=2)^(10)4^n`

Explanation

Now this term is in GP. `16, 64, 256…to 10` terms

∴ Common Ratio `= r = 64/16=4 `

∴ Sum of GP for n terms

= `a(r^n−1)/(r−1) `........(1)

`⇒ a = 16, r = 4, n = 10`

∴ Substituting the above values in (1) we get ⇒ `16(4^(9)−1)/(4−1) ⇒ 16/3(4^9-1)`