Question

Find the sum of n terms of the series 1.4 + 3.7 + 5.10 + …..

Answer 1

Hint: Here, first you have to consider the series 1.4 + 3.7 + 5.10 + ….. as two factors 1, 3, 5,…. and 4, 7, 10, ….. in which both are in AP. Now, we have to find the nth term of both the series and then the sum of the series is calculated as:
${{S}_{n}}=\sum\limits_{n=1}^{n}{\left( 2n-1 \right)\left( 3n+1 \right)}$
By solving the above equation, we will get the value of n.

Complete step-by-step answer:
Here, we have to find the sum of n terms of the series 1.4 +3.7 + 5 .10 +…..
The given series has two factors.
The first factor is of the form 1, 3, 5, … which has a common difference, d = 2. Hence, it forms an AP, with a = 1.
Now, we have to find the term of AP, which is given by the formula:
$\begin{align}
  & {{a}_{n}}=a+(n-1)d \\
 & \Rightarrow {{a}_{n}}=1+(n-1)\times 2 \\
 & \Rightarrow {{a}_{n}}=1+n\times 2-1\times 2 \\
 & \Rightarrow {{a}_{n}}=1+2n-2 \\
 & \Rightarrow {{a}_{n}}=2n-1 \\
\end{align}$
Hence, we got the nth term of the first factor as $2n-1$.Now, consider the second factor which is of the form, 4, 7, 10, …. which has a common difference, d = 3. Therefore, it forms an AP with a = 4.
Now, we have to find the nth term of the AP by the formula:
 $\begin{align}
  & {{a}_{n}}=a+(n-1)d \\
 & \Rightarrow {{a}_{n}}=4+(n-1)\times 3 \\
 & \Rightarrow {{a}_{n}}=4+n\times 3-1\times 3 \\
 & \Rightarrow {{a}_{n}}=4+3n-3 \\
 & \Rightarrow {{a}_{n}}=3n+1 \\
\end{align}$
Hence, the nth term of the given series = $\left( 2n-1 \right)\left( 3n+1 \right)$
Therefore, the sum of the nth term of the given series is given by,
${{S}_{n}}=\sum\limits_{n=1}^{n}{\left( 2n-1 \right)\left( 3n+1 \right)}$
Now by multiplying $2n-1$ by $3n+1$ we obtain:
$\begin{align}
  & {{S}_{n}}=\sum\limits_{n=1}^{n}{2n\times 3n+2n\times 1-1\times 3n-1\times 1} \\
 & {{S}_{n}}=\sum\limits_{n=1}^{n}{6{{n}^{2}}+2n-3n-1} \\
 & {{S}_{n}}=\sum\limits_{n=1}^{n}{6{{n}^{2}}-n-1} \\
\end{align}$
Next, by splitting the terms,
$\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{6{{n}^{2}}}+\sum\limits_{n=1}^{n}{-n}+\sum\limits_{n=1}^{n}{-1}$
In the next step, we can take the minus sign outside of the summation,
$\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{6{{n}^{2}}}-\sum\limits_{n=1}^{n}{n}-\sum\limits_{n=1}^{n}{1}$
Now, since 6 is a constant we can take it, outside of the summation. Hence, we obtain:
${{S}_{n}}=6\sum\limits_{n=1}^{n}{{{n}^{2}}}-\sum\limits_{n=1}^{n}{n}-\sum\limits_{n=1}^{n}{1}$ ….. (1)
We know that that the sum of the squares of first n natural numbers is given by:
 $\sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n(n+1)(2n+1)}{6}$
The sum of the first n natural numbers is given by,
$\sum\limits_{n=1}^{n}{n}=\dfrac{n(n+1)}{2}$
$\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{n}{1}=1+1+1+.....+1 \\
 & \Rightarrow \sum\limits_{n=1}^{n}{1}=n \\
\end{align}$
Now, by substituting all these values in equation (1), we get,
$\Rightarrow {{S}_{n}}=6\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}-n$
Next, by cancellation,
$\Rightarrow {{S}_{n}}=n(n+1)(2n+1)-\dfrac{n(n+1)}{2}-n$
Since, n is the common factor, we can take it outside. Hence, we get it as,
$\Rightarrow {{S}_{n}}=n\left[ (n+1)(2n+1)-\dfrac{(n+1)}{2}-1 \right]$
By simplification we will get:
$\begin{align}
  & \Rightarrow {{S}_{n}}=n\left[ n\times 2n+n\times 1+1\times 2n+1\times 1-\dfrac{n+1}{2}-1 \right] \\
 & \Rightarrow {{S}_{n}}=n\left[ 2{{n}^{2}}+n+2n+1-\dfrac{n+1}{2}-1 \right] \\
 & \Rightarrow {{S}_{n}}=n\left[ 2{{n}^{2}}+3n-\dfrac{n+1}{2} \right] \\
\end{align}$
Next, by taking the LCM we will get:
$\begin{align}
  & \Rightarrow {{S}_{n}}=n\left[ \dfrac{4{{n}^{2}}+6n-(n+1)}{2} \right] \\
 & \Rightarrow {{S}_{n}}=n\left[ \dfrac{4{{n}^{2}}+6n-n-1}{2} \right] \\
 & \Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 4{{n}^{2}}+5n-1 \right] \\
\end{align}$
Hence, the sum of the n terms of the series 1.4 +3.7 + 5 .10 +….. = $\dfrac{n}{2}\left[ 4{{n}^{2}}+5n-1 \right]$.
Note: Here, since there are n terms, so while finding the sum you have to take the summation. There are ‘n’ in the terms and while taking the summation you have to take its limit from 1 to n. Here, you should also know the sum of the first n natural numbers and sum of the squares of the first n natural numbers.