Question

Find the sum of series 1 + (2)(3) + (4) + (5)(6) + 7 + ….. upto 50 terms.

Answer 1

Hint: To solve this question, we will separately consider the sum of series 1 + 4 + 7 + …. using AP, where \[{{a}_{n}}\] is the last term and a is the first term and d is the common difference and finally using the formula, \[{{a}_{n}}=a+\left( n-1 \right)d.\] And sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right].\] Then separately we will calculate the sum of (2)(3) + (4)(5) + …. and add them finally.

Complete step-by-step answer:
We have to find the sum of the series 1 + (2)(3) + (4) + (5)(6) + 7 + ….. upto 50 terms.
Let us represent the sum as S.
\[\Rightarrow S=1+\left( 2 \right)\left( 3 \right)+4+\left( 5 \right)\left( 6 \right)+7+.....50\]
Let us write the terms having one element and the term having two elements separately, then the given series become
\[S=1+4+7+10+13+.....+\left( 2 \right)\left( 3 \right)+\left( 5 \right)\left( 6 \right)+\left( 8 \right)\left( 9 \right)+......\]
Now we need the last term of the series 1 + 4 + 7 + 10 + …..
As we need the terms with a difference of 2 means after 1 leaving 2, 3, we have 4. After 4, leave 5, 6 we have 7. Similarly, all else. When the last term in the series is given as 1 + 4 + 7 + 10 + 13 + …. will be calculated by using the AP (Arithmetic Progression). Let us calculate the last term of the series given by 1 + 4 + 7 + 10 + 13 + ….
It is an Arithmetic progression with the first term, a = 1 and the common difference is 4 – 1 = 3 and the value of n would be half of 50 as there are half of the total 50 terms given in S.
\[\Rightarrow n=\dfrac{50}{2}=25\]
So, we will find the \[{{n}^{th}}\] term now. Using the formula \[{{a}_{n}}=a+\left( n-1 \right)d\] and substituting all the values, we get,
\[\Rightarrow {{a}_{n}}=a+\left( n-1 \right)d\]
\[\Rightarrow {{a}_{n}}=1+\left( 25-1 \right)3\]
\[\Rightarrow {{a}_{n}}=1+24\times 3\]
\[\Rightarrow {{a}_{n}}=1+72\]
\[\Rightarrow {{a}_{n}}=73\]
So, the last term of the series is 1 + 4 + 7 + 10 + 13 ….. is 73.
Finally, we will find the sum of this AP using the formula. The sum will be \[{{S}_{n}}.\]
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
Here, n = 25 and the sum is
\[{{S}_{n}}=\dfrac{25}{2}\left[ 2\times 1+\left( 25-1 \right)\times 3 \right]\]
\[\Rightarrow {{S}_{n}}=\dfrac{25}{2}\left[ 2+72 \right]\]
\[\Rightarrow {{S}_{n}}=\dfrac{25}{2}\times 74\]
\[\Rightarrow {{S}_{n}}=25\times 37\]
\[\Rightarrow {{S}_{n}}=925......\left( i \right)\]
Finally, we will calculate the sum of the left series of S given as
\[\left( 2 \right)\left( 3 \right)+\left( 5 \right)\left( 6 \right)+\left( 8 \right)\left( 9 \right)+.....+\left( 74 \right)\left( 75 \right)\]
Clearly, after 73, 74 and 75 comes. So, the last term of the above series would be (74)(75). Series would be given as
\[\sum\limits_{25}^{n=1}{\left( 3n-1 \right)\left( 3n \right)}\]
\[\Rightarrow \sum\limits_{n=1}^{25}{\left( 3n \right)\left( 3n-1 \right)}\]
\[\Rightarrow \sum\limits_{n=1}^{25}{\left( {{3}^{2}}n-3n \right)}\]
\[\Rightarrow \sum\limits_{n=1}^{25}{\left( 9n-3n \right)}.....\left( ii \right)\]
So, finally we have the sum of the given series as 1 + (2)(3) + 4 + (5)(6) + …. as obtained by adding the equation (i) and (ii).
\[\Rightarrow \text{Sum}=925+\sum\limits_{n=1}^{25}{9n}-\sum\limits_{n=1}^{25}{3n}\]
Hence, this is the required sum of the given series.

Note: Observing the series, we see that we didn’t get any progression AP or GP in series (2)(3) + (5)(6) + (8)(9) +…. So, we have solved them without using any formula and hence directly calculating the sum of it.