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Question: If\[{\rm{a,x}}\]are real numbers and\[\left| {\rm{a}} \right| < 1,\left| {\rm{x}} \right| < 1\], the...

Ifa,x{\rm{a,x}}are real numbers and∣a∣<1,∣x∣<1\left| {\rm{a}} \right| < 1,\left| {\rm{x}} \right| < 1, then1+(1+a)x+(1+a+a2)x2+....∞1 + (1 + {\rm{a)x + (1 + a}} + {{\rm{a}}^2}){{\rm{x}}^2} + ....\infty is equal to
A) 1(1−a)(1−ax)\dfrac{1}{{(1 - {\rm{a}})(1 - {\rm{ax}})}}
B) 1(1−a)(1−x)\dfrac{1}{{(1 - {\rm{a}})(1 - {\rm{x}})}}
C) 1(1−x)(1−ax)\dfrac{1}{{(1 - {\rm{x}})(1 - {\rm{ax}})}}

Explanation

Solution

Here in this question we have to simply solve the equation to get the answer. Firstly we have to simplify the equation by multiplying it with the variable and then we have to subtract the new formed equation from the main equation. So then by simplifying the equation we will get the answer.

Complete step by step solution:
The series given in the question is1+(1+a)x+(1+a+a2)x2+....∞1 + (1 + {\rm{a)x + (1 + a}} + {{\rm{a}}^2}){{\rm{x}}^2} + ....\infty
Let S be the sum of the given series therefore
S=1+(1+a)x+(1+a+a2)x2+....∞{\rm{S = }}1 + (1 + {\rm{a)x + (1 + a}} + {{\rm{a}}^2}){{\rm{x}}^2} + ....\infty ……………. (1)
Now we have to simplify the equation by multiplying it with ax{\rm{ax}}on the both sides. So, we get
⇒axS=ax(1+(1+a)x+(1+a+a2)x2+....∞)\Rightarrow {\rm{axS = ax}}\left( {1 + (1 + {\rm{a)x + (1 + a}} + {{\rm{a}}^2}){{\rm{x}}^2} + ....\infty } \right)
Now taking the termax{\rm{ax}} inside the bracket we get
⇒axS=ax+(a+a2)x2+(a+a2+a3)x3+....∞\Rightarrow {\rm{axS = ax}} + ({\rm{a}} + {{\rm{a}}^2}{\rm{)}}{{\rm{x}}^2}{\rm{ + (a}} + {{\rm{a}}^2} + {{\rm{a}}^3}){{\rm{x}}^3} + ....\infty……………. (2)
Now we have to subtract the equation (2) from equation (1), we get
⇒S−axS=(1+(1+a)x+(1+a+a2)x2+....∞)−(ax+(a+a2)x2+(a+a2+a3)x3+....∞)\Rightarrow {\rm{S}} - {\rm{axS = }}\left( {1 + (1 + {\rm{a)x + (1 + a}} + {{\rm{a}}^2}){{\rm{x}}^2} + ....\infty } \right) - \left( {{\rm{ax}} + ({\rm{a}} + {{\rm{a}}^2}{\rm{)}}{{\rm{x}}^2}{\rm{ + (a}} + {{\rm{a}}^2} + {{\rm{a}}^3}){{\rm{x}}^3} + ....\infty } \right)
⇒S−axS=1+x+x2+x3+.....∞\Rightarrow {\rm{S}} - {\rm{axS = 1 + x}} + {{\rm{x}}^2} + {{\rm{x}}^3} + .....\infty
Now taking S common, we get
⇒S(1−ax)=1+x+x2+x3+.....∞\Rightarrow {\rm{S(1}} - {\rm{ax) = 1 + x}} + {{\rm{x}}^2} + {{\rm{x}}^3} + .....\infty
The right side of the equation becomes a GP with x as common ratio and we know that the sum of infinite terms of a GP is1(1−x)\dfrac{1}{{(1 - {\rm{x}})}}. Therefore, we get
⇒S(1−ax)=1(1−x)\Rightarrow {\rm{S(1}} - {\rm{ax) = }}\dfrac{1}{{(1 - {\rm{x}})}}
⇒S=1(1−x)(1−ax)\Rightarrow {\rm{S = }}\dfrac{1}{{(1 - {\rm{x}}){\rm{(1}} - {\rm{ax)}}}}
Hence, 1(1−x)(1−ax)\dfrac{1}{{(1 - {\rm{x}}){\rm{(1}} - {\rm{ax)}}}}is the sum of the given series.

So, option C is the correct option.

Note:
We have to take the multiplicative factor very carefully because taking the wrong multiplicative factor will lead us to the complex equation which is even harder to simplify. In these types of questions we have to simplify the equation according to the options given because an equation can be simplified in many different ways. So we have to simplify as per the answer required. We should know that the sum of infinite terms of a GP series is equal to1(1−x)\dfrac{1}{{(1 - {\rm{x}})}}.
G.P. series is a,ar,ar2,ar3,ar4,.........{\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........
nthterm=arn−1{{\rm{n}}^{{\rm{th}}}}{\rm{term = a}}{{\rm{r}}^{{\rm{n - 1}}}}
Where, a is the first term of the G.P. and r is the common ratio.