Let S = 2x+5x

Solution:

^{2}+9x^{3}+14x^{4}+20x^{5}+… infinity. The coefficient of n^{th}term is n(n+3)/2. The sum S is:
(1)x(2-x)/(1-x)

^{3}(2) (2-x)/(1-x)^{3 }(3) x(2-x)/(1-x)^{2}(4)None of these
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It is a standard model which deals
with manipulation and simplification of equations.

S = 2x+5x

Multiplying on both sides with 'x':

^{2}+9x^{3}+14x^{4}+20x^{5}+…. -----(1)Multiplying on both sides with 'x':

xS = 2x

^{2}+5x^{3}+9x^{4}+14x^{5}+20x^{6}+…. -----(2)
(1)-(2) gives

(1-x)S = 2x+3x

^{2}+4x^{3}+5x^{4}+… ------(3)
Multiplying on both sides with 'x':

x(1-x)S = 2x

x(1-x)S = 2x

^{2}+3x^{3}+4x^{4}+5x^{5}+… ------(4)
(3)-(4) gives

(1-x)(1-x)S = 2x+x

^{2}+x^{3}+x^{4}+…
(1-x)

^{2}S = x+(x+x^{2}+x^{3}+x^{4}+…... infinite terms)
(1-x)

^{2}S = x + (sum of G.P with initial term ‘x’ and common ratio ‘x’)
“Formula:
sum of infinite G.P = a/(1-r), where a is initial term and r is common ratio”

(1-x)

^{2}S = x + x/(1-x) = (x-x^{2}+x)/(1-x) = (2x-x^{2})/(1-x) = x(2-x)/(1-x)
=> S
= x(2-x)/(1-x)

^{3}**Answer (1)**
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