# Sum of \[1 + 2a + 3{a^2} + 4{a^3} + ........................{\text{ to }}n\]terms.

A. \[\dfrac{{1 + \left( {{a^n}} \right)}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{1 + a}}\]

B. \[\dfrac{{1 - 2\left( {{a^n}} \right)}}{{{{\left( {a - 1} \right)}^2}}} + \dfrac{{n{a^n}}}{{1 - 2a}}\]

C. \[\dfrac{{1 - \left( {{a^n}} \right)}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{1 - a}}\]

D. None of these

Answer

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Hint: In the given series the \[nth\] term is \[n{a^{n - 1}}\]. The sum of the series of \[n\]terms in a Geometric Progression (G.P) with first term \[a\] and common ratio \[r\]is given by \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:

Let \[S = 1 + 2a + 3{a^2} + 4{a^3} + ......................... + nth{\text{ term}}\]

Clearly, \[nth\] term is \[n{a^{n - 1}}\]\[ \Rightarrow S = 1 + 2a + 3{a^2} + 4{a^3} + ......................... + \;n{a^{n - 1}}.....................................\left( 1 \right)\]

Multiplying both sides with ‘\[a\]’, we get

\[

\Rightarrow Sa = \left( {1 + 2a + 3{a^2} + 4{a^3} + ......................... + n{a^{n - 1}}} \right)a \\

\Rightarrow Sa = 0 + a + 2{a^2} + 3{a^3} + 4{a^4} + ......................... + \left( {n - 1} \right){a^{n - 1}} + n{a^n}...........................................\left( 2 \right) \\

\]

Subtracting equation (2) from (1), we get

\[

\Rightarrow S - Sa = \left( {1 + 2a + 3{a^2} + 4{a^3} + ......................... + \;n{a^{n - 1}}} \right) - \left( {0 + a + 2{a^2} + 3{a^3} + 4{a^4} + ......................... + n{a^n}} \right) \\

\Rightarrow S\left( {1 - a} \right) = 1 + a + {a^2} + {a^3} + .............................. + {a^{n - 1}} - n{a^n} \\

\]

Clearly, the above series is in G.P of\[n\]terms with first term 1 and common ratio \[a\]

We know that, the sum of the series of \[n\]terms in a Geometric Progression (G.P) with first term \[a\] and common ratio \[r\]is given by \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\].

By using the above formula, we have

\[ \Rightarrow S\left( {1 - a} \right) = \dfrac{{1\left( {{a^n} - 1} \right)}}{{\left( {a - 1} \right)}} - n{a^n}\]

Dividing both sides with \[\left( {1 - a} \right)\], we get

\[

\Rightarrow \dfrac{{S\left( {1 - a} \right)}}{{\left( {1 - a} \right)}} = \dfrac{{1\left( {{a^n} - 1} \right)}}{{\left( {a - 1} \right)}} \times \dfrac{1}{{\left( {1 - a} \right)}} - n{a^n} \times \dfrac{1}{{\left( {1 - a} \right)}} \\

\Rightarrow S = \dfrac{{ - 1\left( {1 - {a^n}} \right)}}{{\left( {a - 1} \right)}} \times \dfrac{{ - 1}}{{\left( {a - 1} \right)}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}} \\

\therefore S = \dfrac{{1\left( {1 - {a^n}} \right)}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}} \\

\]

Therefore, the sum of the \[1 + 2a + 3{a^2} + 4{a^3} + ........................{\text{ to }}n\]terms is \[\dfrac{{1 - {a^n}}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}}\]

Thus, the correct option is C. \[\dfrac{{1 - {a^n}}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}}\]

Note: In equation (2) we added 0 as the first term so that you can subtract it from the equation (1) with no confusion. This has been done only for your understanding. Any way by doing this the value of the equation (2) doesn’t change. And observe that \[S\left( {1 - a} \right)\] has \[n + 1\]terms but, we have taken only \[n\]terms for the summation of the series by leaving the last term as it is.

Complete step-by-step answer:

Let \[S = 1 + 2a + 3{a^2} + 4{a^3} + ......................... + nth{\text{ term}}\]

Clearly, \[nth\] term is \[n{a^{n - 1}}\]\[ \Rightarrow S = 1 + 2a + 3{a^2} + 4{a^3} + ......................... + \;n{a^{n - 1}}.....................................\left( 1 \right)\]

Multiplying both sides with ‘\[a\]’, we get

\[

\Rightarrow Sa = \left( {1 + 2a + 3{a^2} + 4{a^3} + ......................... + n{a^{n - 1}}} \right)a \\

\Rightarrow Sa = 0 + a + 2{a^2} + 3{a^3} + 4{a^4} + ......................... + \left( {n - 1} \right){a^{n - 1}} + n{a^n}...........................................\left( 2 \right) \\

\]

Subtracting equation (2) from (1), we get

\[

\Rightarrow S - Sa = \left( {1 + 2a + 3{a^2} + 4{a^3} + ......................... + \;n{a^{n - 1}}} \right) - \left( {0 + a + 2{a^2} + 3{a^3} + 4{a^4} + ......................... + n{a^n}} \right) \\

\Rightarrow S\left( {1 - a} \right) = 1 + a + {a^2} + {a^3} + .............................. + {a^{n - 1}} - n{a^n} \\

\]

Clearly, the above series is in G.P of\[n\]terms with first term 1 and common ratio \[a\]

We know that, the sum of the series of \[n\]terms in a Geometric Progression (G.P) with first term \[a\] and common ratio \[r\]is given by \[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\].

By using the above formula, we have

\[ \Rightarrow S\left( {1 - a} \right) = \dfrac{{1\left( {{a^n} - 1} \right)}}{{\left( {a - 1} \right)}} - n{a^n}\]

Dividing both sides with \[\left( {1 - a} \right)\], we get

\[

\Rightarrow \dfrac{{S\left( {1 - a} \right)}}{{\left( {1 - a} \right)}} = \dfrac{{1\left( {{a^n} - 1} \right)}}{{\left( {a - 1} \right)}} \times \dfrac{1}{{\left( {1 - a} \right)}} - n{a^n} \times \dfrac{1}{{\left( {1 - a} \right)}} \\

\Rightarrow S = \dfrac{{ - 1\left( {1 - {a^n}} \right)}}{{\left( {a - 1} \right)}} \times \dfrac{{ - 1}}{{\left( {a - 1} \right)}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}} \\

\therefore S = \dfrac{{1\left( {1 - {a^n}} \right)}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}} \\

\]

Therefore, the sum of the \[1 + 2a + 3{a^2} + 4{a^3} + ........................{\text{ to }}n\]terms is \[\dfrac{{1 - {a^n}}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}}\]

Thus, the correct option is C. \[\dfrac{{1 - {a^n}}}{{{{\left( {a - 1} \right)}^2}}} - \dfrac{{n{a^n}}}{{\left( {1 - a} \right)}}\]

Note: In equation (2) we added 0 as the first term so that you can subtract it from the equation (1) with no confusion. This has been done only for your understanding. Any way by doing this the value of the equation (2) doesn’t change. And observe that \[S\left( {1 - a} \right)\] has \[n + 1\]terms but, we have taken only \[n\]terms for the summation of the series by leaving the last term as it is.

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