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Question

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


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
(1x)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(1x)s=x+3x2+5x2+7x4+.... ----(4)
Subtracting equation 4 from equation 3, we get
(1x)(1x)s=1+2(x+x2+x3+......)
(1x)2s=1+2(x1+x)=1+x1x

s=1+x(1x)3
Substitute x =15
s=1+15(115)3=7532
32 x 5 = 75

Mathematics

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