If \[|a| < 1\] and \[|b| < 1\] then \[a(a + b) + {a^2}({a^2} + {b^2}) + {a^3}({a^3} + {b^3}) + ..........\] up to \[\infty \] is
A. \[\dfrac{{{b^2}}}{{1 - {b^2}}} + \dfrac{{ab}}{{1 - ab}}\]
B. \[\dfrac{1}{{1 - {a^2}}} + \dfrac{{ab}}{{1 - ab}}\]
C. \[\dfrac{1}{{1 - {a^2}}} + \dfrac{{ab}}{{1 - ab}}\]
D. \[\dfrac{{{a^2}}}{{1 - {a^2}}} + \dfrac{{ab}}{{1 - ab}}\]

Hint: We express the given series in summation forms and we simplify this. After simplifying we again we expand the summation into series form. We obtain a series which are in geometric progression. We know the sum of infinite terms in geometric series is \[{S_\infty } = \dfrac{a}{{(1 - r)}}\] , ‘a’ is the first term and ‘r’ is the common ratio.

Complete step-by-step answer:
We have \[a(a + b) + {a^2}({a^2} + {b^2}) + {a^3}({a^3} + {b^3}) + ..........\infty {\text{ - - - - - - - (1)}}\]
Now we can express series (1) in summation form that is,
 \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {{a^n}({a^n} + {b^n})} \]
Now multiplying with in the summation,
 \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {({a^n}.{a^n} + {a^n}{b^n})} \]
We know \[{a^n}.{a^m} = {a^{n + m}}\] , we get:
 \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {({a^{n + n}} + {a^n}{b^n})} \]
We know \[{a^n}{b^n} = {(ab)^n}\] , we get:
 \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {\left( {{a^{2n}} + {{\left( {ab} \right)}^n}} \right)} \]
Expanding the summation, we have:
 \[ \Rightarrow S = \sum\limits_{n = 1}^\infty {\left( {{a^{2n}}} \right)} + \sum\limits_{n = 1}^\infty {{{\left( {ab} \right)}^n}} {\text{ - - - - - - (2)}}\]
Now we express the each summation in series form we have,
Now, \[\sum\limits_{n = 1}^\infty {\left( {{a^{2n}}} \right)} = \left( {{a^2} + {a^4} + {a^6} + ........ + \infty } \right)\]
We can see that the series are in geometric progression,
Common ratio \[r = {a^2}\] and first term is \[a = {a^2}\] .
Now the sum of infinite terms in geometric series is \[{S_\infty } = \dfrac{a}{{(1 - r)}}\]
Then \[ \Rightarrow \sum\limits_{n = 1}^\infty {\left( {{a^{2n}}} \right)} = {S_\infty } = \dfrac{{{a^2}}}{{1 - {a^2}}}{\text{ - - - - - - - (3)}}\]
Now, \[\sum\limits_{n = 1}^\infty {{{\left( {ab} \right)}^n}} = \left( {ab + {{(ab)}^2} + {{(ab)}^3} + ........ + \infty } \right)\]
Similarly, this series is also in geometric progression,
Common ratio \[r = ab\] and first term is \[a = ab\] .
Now the sum of infinite terms in geometric series is \[{S_\infty } = \dfrac{a}{{(1 - r)}}\]
 \[ \Rightarrow \sum\limits_{n = 1}^\infty {{{\left( {ab} \right)}^n}} = {S_\infty } = \dfrac{{ab}}{{1 - ab}}{\text{ - - - - - (4)}}\]
Now substituting equation (3) and (4) in equation (2) we have,
 \[ \Rightarrow S = \dfrac{{{a^2}}}{{1 - {a^2}}} + \dfrac{{ab}}{{1 - ab}}\]
So, the correct answer is “Option D”.

Note: In geometric progression the common ratio is given by \[r = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\] . In arithmetic progression we take the common difference of adjacent pairs \[d = {a_n} - {a_{n - 1}}\] . In the given series if we have common differences then it is in arithmetic progression. If the given series have a common ratio then it is in geometric progression. Sum of ‘n’ terms in arithmetic progression is \[{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right] \] .