1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If S=1+22x+32x2+42x3+......∞ where x=15. Find the value of 32 S.

Open in App
Solution

## This series is not AP, GP, HP and AGP How can we proceed? But we see 12,22,32,42..........∞ are square numbers and 1,x,x2,x3,x4,........∞ is a geometric progression. If we subtract square numbers. We can form an AP. To subtract square numbers multiply given series by common ratio and then subtract it with original series. s=1+22x+32x2+42x3+......∞ -----(1) Multiply equation 1 by common ratio x xs=x+22x2+32x3+.....∞ -----(2) Subtract equation 2 from equation 1 (1−x)s=1+3x+5x2+7x3+....∞ -----(3) Now, we will see we got Arithmetic - geometric progression. At this stage you should be able to solve a AGP Again multiply equation 3 by common ratio x x(1−x)s=x+3x2+5x2+7x4+....∞ ----(4) Subtracting equation 4 from equation 3, we get (1−x)(1−x)s=1+2(x+x2+x3+......∞) (1−x)2s=1+2(x1+x)=1+x1−x s=1+x(1−x)3 Substitute x =15 s=1+15(1−15)3=7532 32 x 5 = 75

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Properties of GP
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program