Question

Find the 5th term and the sum of the first 5 terms of the geometric sequence 2, -1/2,1/8,....

Determine the first terms of the terms of the sequence of {(-1/n)power to n} and give their corresponding sum.

Answer 1

q=b_n+1/b=-0.5/2=-0.25

b₅₌b₁q^n-1=2x(-0.25)^5-1=2/256=1/125

$S_n=\frac{b_1(1-q^n)}{q-1}=\frac{2(1-(-0.25)^5}{(-0.25)-1} =\frac{2(1+1/1024)}{-5/4}=\frac{8(1+1/1024)}{-5}=-(8/5 +1/128) =- 1.61$

First terms of (-1/n)ⁿ are: -1, 1/4, -1/27

$q=\frac{1/4}{-1}=-1/4$

$S_3=\frac{-1(1-(-1/4)^3)}{(-0.25-1)}=\frac{-65/64}{-5/4}=\frac{65x4}{64x5}=\frac{13}{16}$