Question

Find the sum of the series : 1.n + 2.(n – 1) + 3.(n – 2) + ….. + (n – 1).2 + n.1.

Answer 1

Hint:- We can find the general value of any term r i.e. ${T_r}$. And then we can apply summation to ${T_r}$ and find the sum from r = 1 to n because the total number of terms is n.

Complete step-by-step answer:
Let the ${r^{th}}$ term be ${T_r}$ .
As we can see from the given series that each term consists of two numbers (let x and y). And the resultant term is the product of both numbers (i.e. x.y).
The first number (i.e. x) of the first term of the series is 1 and the second number (i.e. y) of the first term is n.
Now for each term the first number increases by 1 and the second number decreases by 1.
And the first number of the last term of the series is n and the second number of the last term of the series is 1.
So, the total number of terms in the series will be n.
And the ${r^{th}}$ term will be $r\left( {n - r + 1} \right)$
So, ${T_r} = r\left( {n - r + 1} \right) = rn - {r^2} + r$
Now as we know that we had to find the sum of n terms of the series whose ${r^{th}}$ term is $rn - {r^2} + r$.
So, let ${S_n}$ be the sum of n terms of the series.
So, applying summation from r = 1 to r = n in ${T_r}$ to find the sum of the series.
${S_n} = \sum\limits_{r = 1}^n {{T_r}} = \sum\limits_{r = 1}^n {rn - {r^2} + r} = n\sum\limits_{r = 1}^n r - \sum\limits_{r = 1}^n {{r^2}} + \sum\limits_{r = 1}^n r = \left( {n + 1} \right)\sum\limits_{r = 1}^n r - \sum\limits_{r = 1}^n {{r^2}}$ (1)
Now as we know that the sum of the squares of the first n natural numbers (i.e. 1 to n) is equal to $\sum\limits_{r = 1}^n {{r^2}}$.
And, $\sum\limits_{r = 1}^n {{r^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ (2)
And the sum of first n natural numbers (i.e. 1 to n) is equal to $\sum\limits_{r = 1}^n r$.
And, $\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$ (3)
So, now putting the value of equation 2 and 3 in equation 1.
${S_n} = \left( {n + 1} \right)\dfrac{{n\left( {n + 1} \right)}}{2} - \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\left( {n + 1} \right) - \dfrac{{\left( {2n + 1} \right)}}{3}} \right]$
${S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{n + 2}}{3}} \right] = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$
Hence, the sum of the series 1.n + 2.(n – 1) + 3.(n – 2) + ….. + (n – 1).2 + n.1 will be $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$.

Note:- Whenever we come up with this type of problem then we had to first find the number of terms in the series (like here n) and then we had to find the ${r^{th}}$ term of the series by observing the each term that how value of each term depends on its count (i.e. r). And after that we can apply a summation formula to the ${r^{th}}$ term to find the sum of series (i.e. sum of terms from first term to the last term (n)). And after that we had to use the formula of sum of first n numbers i.e. $\dfrac{{n\left( {n + 1} \right)}}{2}$ and formula of sum of square of first n numbers i.e. $\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$ to find the value of summation. This will be the easiest and efficient way to find the solution of the problem.