Question

Find the ${{n}^{th}}$ term and the sum to n terms of the following series $2+7x+25{{x}^{2}}+91{{x}^{3}}+.........$

Answer 1

Hint: Break the terms of the given series as the sum of exponents of 3 and 4, that is, \[\left( {{3}^{0}}+{{4}^{0}} \right)+\left( {{3}^{1}}+{{4}^{1}} \right)x+\left( {{3}^{2}}+{{4}^{2}} \right){{x}^{2}}+\left( {{3}^{3}}+{{4}^{3}} \right){{x}^{3}}+......\]. Start from the power ‘0’ of 3 and 4 and use the formula: ${{T}_{n}}=a{{r}^{n-1}}$ to determine the ${{n}^{th}}$ term of the G.P series obtained. Here, ${{T}_{n}}$ is the ${{n}^{th}}$ term, ‘a’ is the first term and ‘r’ is the common ratio of the series. To determine the sum of these terms use the relation: ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$.

Complete step-by-step answer:
In this question geometric progression will be used. In Mathematics, a G.P. also known as geometric sequence is a sequence of numbers where each term after its predecessor is obtained by multiplying the previous term with a fixed non-zero number known as the common ratio of the G.P.
For example: \[2,\text{ }4,\text{ }8,\text{ }16,\text{ }32,\text{ }.........\] is a G.P. with common ratio $2$.
Now, we come to the question. We have been given the series $2+7x+25{{x}^{2}}+91{{x}^{3}}+.........$
This can be written as:
\[\begin{align}
  & \left( {{3}^{0}}+{{4}^{0}} \right)+\left( {{3}^{1}}+{{4}^{1}} \right)x+\left( {{3}^{2}}+{{4}^{2}} \right){{x}^{2}}+\left( {{3}^{3}}+{{4}^{3}} \right){{x}^{3}}+...... \\
 & =\left( {{3}^{0}}{{x}^{0}}+{{3}^{1}}{{x}^{1}}+{{3}^{2}}{{x}^{2}}+{{3}^{3}}{{x}^{3}}+..... \right)+\left( {{4}^{0}}{{x}^{0}}+{{4}^{1}}{{x}^{1}}+{{4}^{2}}{{x}^{2}}+{{4}^{3}}{{x}^{3}}+..... \right) \\
 & =\left( 1+{{3}^{1}}{{x}^{1}}+{{3}^{2}}{{x}^{2}}+{{3}^{3}}{{x}^{3}}+..... \right)+\left( 1+{{4}^{1}}{{x}^{1}}+{{4}^{2}}{{x}^{2}}+{{4}^{3}}{{x}^{3}}+..... \right) \\
\end{align}\]
Now, the pattern is clear to us. Clearly, we can see that for \[\left( 1+{{3}^{1}}{{x}^{1}}+{{3}^{2}}{{x}^{2}}+{{3}^{3}}{{x}^{3}}+..... \right)\],
a = first term = 1 and r = common ratio = 3x.
Now, for \[\left( 1+{{4}^{1}}{{x}^{1}}+{{4}^{2}}{{x}^{2}}+{{4}^{3}}{{x}^{3}}+..... \right)\],
a = first term = 1 and r = common ratio = 4x.
Therefore, applying the formula for ${{n}^{th}}$ term of the G.P, we get,
$\begin{align}
  & {{T}_{n}}=a{{r}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=1\times {{\left( 3x \right)}^{n-1}}+1\times {{\left( 4x \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}={{\left( 3x \right)}^{n-1}}+{{\left( 4x \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}={{\left( x \right)}^{n-1}}\left( {{3}^{n-1}}+{{4}^{n-1}} \right) \\
\end{align}$
We know that sum of n terms of a G.P is ${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$. Therefore, the sum of the given series is:
$\begin{align}
  & {{S}_{n}}=\dfrac{1\times \left( {{3}^{n}}{{x}^{n}}-1 \right)}{3x-1}+\dfrac{1\times \left( {{4}^{n}}{{x}^{n}}-1 \right)}{4x-1} \\
 & \therefore {{S}_{n}}\text{ }=\dfrac{{{\left( 3x \right)}^{n}}-1}{3x-1}+\dfrac{{{\left( 4x \right)}^{n}}-1}{4x-1} \\
\end{align}$

Note:We can check our answer by substituting the value of ‘n’ in the expression of sum of n terms. Also, as you can see that we have broken the terms to extract a pattern, otherwise it would have been very difficult to solve the problem.