Question

The sum of the series $1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $ is given by
A) ${n^2} + 1$
B) $n(n + 1)$
C) $n{\left( {1 + \dfrac{1}{n}} \right)^2}$
D) ${n^2}$

Answer 1

Hint: For an infinite series ${a_1} + {a_2} + {a_3} + ... + \infty $, a quantity ${S_n} = {a_1} + {a_2} + {a_3} + ... + {a_n}$, which involves adding only the first n terms, is called a partial sum of the series. If ${S_n}$ approaches a fixed number S as n becomes larger and larger, the series is said to converge. In this case, S is called the sum of the series.
First, multiply infinite series with $\left( {1 + \dfrac{1}{n}} \right)$. Then subtract both the series. First, simplify the left-hand side and then the right-hand side. Solve the equations with the help of LCM. Then Combine both sides and we will get the final answer. We will use the formula as stated below:
$1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty = \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}$

Complete step-by-step solution:
The given infinite series is,
$ \Rightarrow {S_n} = 1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $
Let us multiply $\left( {1 + \dfrac{1}{n}} \right)$ both sides.
Therefore,
$ \Rightarrow \left( {1 + \dfrac{1}{n}} \right){S_n} = \left( {1 + \dfrac{1}{n}} \right) + 2{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $
Now, we will subtract ${S_n}$ and$\left( {1 + \dfrac{1}{n}} \right){S_n}$ .
Hence, the left-hand side is.
$ \Rightarrow {S_n} - \left( {1 + \dfrac{1}{n}} \right){S_n}$
Solve the above equation.
$ \Rightarrow {S_n} - {S_n} - \left( {\dfrac{1}{n}} \right){S_n}$
So, we get.
$ \Rightarrow - \left( {\dfrac{1}{n}} \right){S_n}$
Now, we will subtract $1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $ and$\left( {1 + \dfrac{1}{n}} \right) + 2{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $ .
Hence, the right-hand side is.
$ \Rightarrow \left[ {1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{{\left( {1 + \dfrac{1}{n}} \right)}^2} + ... + \infty } \right] - \left[ {\left( {1 + \dfrac{1}{n}} \right) + 2{{\left( {1 + \dfrac{1}{n}} \right)}^2} + ... + \infty } \right]$
Let us subtract the above step.
 $ \Rightarrow 1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty - \left( {1 + \dfrac{1}{n}} \right) - 2{\left( {1 + \dfrac{1}{n}} \right)^2} - ... - \infty $
Therefore, the answer will be.
$ \Rightarrow 1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $
As we already know, the value of $1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty $ is $\dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}$
So, we put that value.
$ \Rightarrow \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}$
Let us simplify.
$ \Rightarrow \dfrac{1}{{1 - \left( {\dfrac{{n + 1}}{n}} \right)}}$
Let us take LCM.
$ \Rightarrow \dfrac{1}{{\left( {\dfrac{{n - n - 1}}{n}} \right)}}$
So, we will get it.
$ \Rightarrow \dfrac{1}{{\left( {\dfrac{{ - 1}}{n}} \right)}}$
Therefore,
$ \Rightarrow - n$
Now, let us combine the left-hand side and right-hand side.
So, we will get it.
$ \Rightarrow - \left( {\dfrac{1}{n}} \right){S_n} = - n$
Multiply both sides by$ - n$ .
$ \Rightarrow {S_n} = {n^2}$
Hence, the final answer is ${n^2}$ .

Option D is the correct answer.

Note: The simplest way to express a result is as a series. To get the numerical value only we need to compute the value. For that, we have to remember the infinite series formula. Here, we use the formula$1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty = \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}$. Make sure that the LCM of two or more numbers is greater than or equal to the greatest number of given numbers.