Q.116.Sum of a series

Question 116.
Find the sum:
(2 + 6)/4^100 + (2 + 2*6)/4^99 + (2 + 3*6)/4^98 + ... + (2 + 99*6)/4^2 + (2 + 100*6)/4.

Answer 116.
(2 + 6)/4^100 + (2 + 2*6)/4^99 + (2 + 3*6)/4^98 + ... + (2 + 99*6)/4^2 + (2 + 100*6)/4

= (2/4^100)(1 + 4 + 4^2 + 4^3 + ... + 4^99) + (6/4^100)(1 + 2*4 + 3*4^2 + ... + 100*4^99)
= S1 + S2
 S1
= (2/4^100)(4^100 - 1)/(4 - 1)
= (2/4^100) (4^100 - 1)/3

For S2, we note that
1 + x + x^2 + x^3 + ... + x^100 = (x^101 - 1)/(x - 1)
Diffrentiating w.r.t. x,
1 + 2x + 3x^2 + ... + 100x^99 = [(x - 1)*101x^100 - (x^101 - 1)]/(x - 1)^2
Plugging x = 4,
1 + 2*4 + 3*4^2 + ... + 100*4^99 = [3*(101)*4^100 - 4^101 + 1]/9
=> S2
= (6/4^100) * [3*(101)*4^100 - 4^101 + 1]/9
= (2/4^100) * (299 * 4^100 + 1)/3

=> S1 + S2
= (2/4^100) [(4^100 - 1) + (299 * 4^100 + 1)]/3
= 200.

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