Question

Find the sum of the series $1+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........$ up to 11 terms?
(a) 915
(b) 946
(c) 945
(d) 916

Answer 1

Hint: We start solving the problem by finding the general equation to represent each term of the series. After finding the general equation, we take summation of it up to n terms. Once we find the summation up to n terms, we substitute 11 in place in n and make required calculations to get the desired result.

Complete step-by-step answer:
According to the problem, we need to find the sum of the series $1+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........$ up to 11 terms.
Let us find the general term of the series to solve for the sum of the series.
$\Rightarrow \left( 1\times 1 \right)+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........$ .
$\Rightarrow \left( 1\times \left( 2-1 \right) \right)+\left( 2\times \left( 4-1 \right) \right)+\left( 3\times \left( 6-1 \right) \right)+\left( 4\times \left( 8-1 \right) \right)........$ .
$\Rightarrow \left( 1\times \left( 2\left( 1 \right)-1 \right) \right)+\left( 2\times \left( 2\left( 2 \right)-1 \right) \right)+\left( 3\times \left( 2\left( 3 \right)-1 \right) \right)+\left( 4\times \left( 2\left( 4 \right)-1 \right) \right)........$ ---(1).
We can see that each term is of the form $r\times \left( 2r-1 \right)$ for $r=1,2,3,......n$.
So, we get general term as $r\times \left( 2r-1 \right)=\left( 2{{r}^{2}}-r \right)$.
We can represent sum of the series in equation (1) as $\sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=\sum\limits_{r=1}^{n}{2{{r}^{2}}}-\sum\limits_{r=1}^{n}{r}$.
$\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=2\sum\limits_{r=1}^{n}{{{r}^{2}}}-\sum\limits_{r=1}^{n}{r}$.
We know that sum of the squares of the first n natural numbers is $\sum\limits_{r=1}^{n}{{{r}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$ and sum of the first n natural numbers is $\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}$.
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=2\left( \dfrac{n\times \left( n+1 \right)\times \left( 2n+1 \right)}{6} \right)-\left( \dfrac{n\left( n+1 \right)}{2} \right)\].
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=\left( \dfrac{n\times \left( n+1 \right)\times \left( 2n+1 \right)}{3} \right)-\left( \dfrac{n\left( n+1 \right)}{2} \right)\].
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=n\times \left( n+1 \right)\times \left( \dfrac{\left( 2n+1 \right)}{3}-\dfrac{1}{2} \right)\].
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=n\times \left( n+1 \right)\times \left( \dfrac{2\times \left( 2n+1 \right)-3}{6} \right)\].
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=n\times \left( n+1 \right)\times \left( \dfrac{4n+2-3}{6} \right)\].
\[\Rightarrow \sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}=\dfrac{n\times \left( n+1 \right)\times \left( 4n-1 \right)}{6}\] ---(2).
Now we substitute 11 in place of n in equation (2).
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=\dfrac{11\times \left( 11+1 \right)\times \left( 4\left( 11 \right)-1 \right)}{6}\].
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=\dfrac{11\times \left( 12 \right)\times \left( 44-1 \right)}{6}\].
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=11\times 2\times 43\].
\[\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=946\].
We have found the sum of the series $1+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........$ up to 11 terms as 946.
∴ The sum of the series $1+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........$ up to 11 terms as 946.
The correct option for the given problem is (b).

Note: Whenever we get this type of problem, we try to find the general equation of the terms which makes our sum easier. We can also take 10 in place of n while finding the general summation for n terms. We should not make any mistakes while calculating the general equation, sum and others. Similarly, we expect problems involving sum of cubes of first n natural numbers.