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Question: Find the sum of the series \(1+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \r...

Find the sum of the series 1+(2×3)+(3×5)+(4×7)........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

Explanation

Solution

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+(2×3)+(3×5)+(4×7)........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.
⇒(1×1)+(2×3)+(3×5)+(4×7)........\Rightarrow \left( 1\times 1 \right)+\left( 2\times 3 \right)+\left( 3\times 5 \right)+\left( 4\times 7 \right)........ .
⇒(1×(2−1))+(2×(4−1))+(3×(6−1))+(4×(8−1))........\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)........ .
⇒(1×(2(1)−1))+(2×(2(2)−1))+(3×(2(3)−1))+(4×(2(4)−1))........\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×(2r−1)r\times \left( 2r-1 \right) for r=1,2,3,......nr=1,2,3,......n.
So, we get general term as r×(2r−1)=(2r2−r)r\times \left( 2r-1 \right)=\left( 2{{r}^{2}}-r \right).
We can represent sum of the series in equation (1) as ∑r=1n(2r2−r)\sum\limits_{r=1}^{n}{\left( 2{{r}^{2}}-r \right)}.
⇒∑r=1n(2r2−r)=∑r=1n2r2−∑r=1nr\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}.
⇒∑r=1n(2r2−r)=2∑r=1nr2−∑r=1nr\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 ∑r=1nr2=n(n+1)(2n+1)6\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 ∑r=1nr=n(n+1)2\sum\limits_{r=1}^{n}{r}=\dfrac{n\left( n+1 \right)}{2}.
⇒∑r=1n(2r2−r)=2(n×(n+1)×(2n+1)6)−(n(n+1)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).
⇒∑r=1n(2r2−r)=(n×(n+1)×(2n+1)3)−(n(n+1)2)\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).
⇒∑r=1n(2r2−r)=n×(n+1)×((2n+1)3−12)\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).
⇒∑r=1n(2r2−r)=n×(n+1)×(2×(2n+1)−36)\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).
⇒∑r=1n(2r2−r)=n×(n+1)×(4n+2−36)\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).
⇒∑r=1n(2r2−r)=n×(n+1)×(4n−1)6\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).
⇒∑r=110(2r2−r)=11×(11+1)×(4(11)−1)6\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}.
⇒∑r=110(2r2−r)=11×(12)×(44−1)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}.
⇒∑r=110(2r2−r)=11×2×43\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=11\times 2\times 43.
⇒∑r=110(2r2−r)=946\Rightarrow \sum\limits_{r=1}^{10}{\left( 2{{r}^{2}}-r \right)}=946.
We have found the sum of the series 1+(2×3)+(3×5)+(4×7)........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+(2×3)+(3×5)+(4×7)........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.