Question

# If the sum to infinity of the series $$1+4x+7x^2+10x^3+\cdots$$ is $$\dfrac{35}{16}$$, then $$x=$$

A
15
B
25
C
37
D
17

Solution

## The correct option is A $$\displaystyle \frac{1}{5}$$Let$${ S }_{ \infty }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty$$      $$...(1)$$Now, multiply by $$x$$ throughout in eqution $$(1)$$; we get$$x{ S }_{ \infty }=x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty$$      $$...(2)$$Subtracting $$(2)$$ from $$(1)$$; we get$${S}_{ \infty }-x{ S }_{ \infty }=(1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty )-(x+4{ x }^{ 2 }+7{ x }^{ 3 }+10{ x }^{ 4 }+...\infty )$$$$\Rightarrow(1-x){ S }_{ \infty }=1+4x+7{ x }^{ 2 }+10{ x }^{ 3 }+...\infty -x-4{ x}^{ 2 }-7{ x }^{ 3 }-10{ x }^{ 4 }-...\infty$$$$\Rightarrow (1-x){ S }_{ \infty }=1+3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty$$Notice that the series $$3x+3{ x }^{ 2 }+3{ x }^{ 3 }+...\infty$$ is geometric series with the first term $$a=3x$$ and the common ratio $$r=x$$.Now, use the formula for the sum of an infinite geometric series.$$\Rightarrow (1-x){ S }_{ \infty }=1+\dfrac { 3x }{ (1-x) }$$, for$$\left| x \right| <1$$$$\Rightarrow (1-x){ S }_{ \infty }=\dfrac { 1+2x }{ (1-x) }$$, for$$\left| x \right| <1$$Given that, $${ S }_{ \infty }=\dfrac { 35 }{ 16 }$$, substitute for $${ S }_{ \infty }$$ in the above equation; we get$$(1-x)\dfrac { 35 }{ 16 } =\dfrac { 1+2x }{ (1-x) }$$$$\Rightarrow35{ (1-x) }^{ 2 }=16(1+2x)$$$$\Rightarrow 35(1-2x+{ x }^{ 2 })=16+32x$$$$\Rightarrow 35{ x }^{ 2 }-102x+19=0$$$$\Rightarrow (7x-19)(5x-1)=0$$$$\Rightarrow x=\dfrac { 19 }{ 7 }$$ or $$x=\dfrac { 1 }{ 5 }$$But $$x\neq \dfrac { 19 }{ 7 }$$, because for infinity series, $$\left| x \right| <1$$.Therefore, $$x=\dfrac { 1 }{ 5 }$$.Maths

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