Question

The sum of n terms in the ${{n}^{th}}$ bracket of the series (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + …. Is

Answer 1

Hint: Now we will first form a series of just the first term of the brackets. Now again we will add 0 to this series and form another series. Now we will subtract the two obtained series to find ${{t}_{n}}$ . Now we want to solve an AP. We know that the sum of AP is given by $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . Hence we can easily find the value of ${{t}_{n}}$ . Now note that ${{t}_{n}}$ is the first term of the ${{n}^{th}}$ bracket. And the ${{n}^{th}}$ bracket will have $2n-1$ terms. Hence using the sum of AP we can again find the sum of all the terms in the ${{n}^{th}}$ bracket.

Complete step by step answer:
Now consider the given series (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9)
Now we will form a series by considering just the first number of the brackets.
Hence we get $1+2+5+10.....+{{t}_{n}}$
${{S}_{n}}=1+2+5+10.....+{{t}_{n}}........................\left( 1 \right)$
${{S}_{n}}=0+1+2+5+10.....{{t}_{n-1}}+{{t}_{n}}....................\left( 2 \right)$
Now subtracting equation (2) from equation (1) we get.
$\begin{align}
  & 0=1+1+3+5+7+....\left( n-1 \right)terms-{{t}_{n}} \\
 & \Rightarrow {{t}_{n}}=1+\left( 1+3+5+7+....\left( n-1 \right)terms \right)......................\left( 3 \right) \\
\end{align}$
Now consider the series 1 + 3 + 5 + 7 ….. (n-1) terms.
This is an AP with a = 1 and d = 2.
Now we know that sum of n terms AP is given by $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Hence sum of n – 1 terms is given by $\dfrac{\left( n-1 \right)}{2}\left[ 2a+\left( n-1-1 \right)d \right]$
Hence sum of n – 1 terms are $\dfrac{\left( n-1 \right)}{2}\left[ 2a+\left( n-2 \right)d \right]$
Now we know that a = 1 and d = 2.
Hence we have sum of n – 1 terms are $\dfrac{\left( n-1 \right)}{2}\left[ 2+\left( n-2 \right)2 \right]$
Hence we have 1 + 3 + 5 + 7 ….. (n-1) terms $=\dfrac{\left( n-1 \right)}{2}\left[ 2n-2 \right]={{\left( n-1 \right)}^{2}}$
Now from equation (3) we get,
Hence we get ${{t}_{n}}=1+1+3+5+7+...\left( n-1 \right)terms=1+{{\left( n-1 \right)}^{2}}$
${{t}_{n}}=1+{{\left( n-1 \right)}^{2}}..........................\left( 4 \right)$
Now we have the first term of ${{n}^{th}}$ bracket.
Now we know that in first bracket we have 1 term.
In second bracket we have 3 terms.
And so on in ${{n}^{th}}$ bracket we have $2n-1$ terms.
Hence we have that the ${{n}^{th}}$ bracket is an AP with a = $1+{{\left( n-1 \right)}^{2}}$ d = 1.
So sum of n terms is $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ .
Hence we have sum of $2n-1$ is $\dfrac{2n-1}{2}\left[ 2a+\left( 2n-1-1 \right)d \right]$
Now let us substitute a = $1+{{\left( n-1 \right)}^{2}}$ and d = 1 in the equation.
\[\begin{align}
  & S=\dfrac{2n-1}{2}\left[ 2\left( 1+{{\left( n-1 \right)}^{2}} \right)+\left( 2n-2 \right) \right] \\
 & \Rightarrow S=\left( 2n-1 \right)\left[ \left( 1+{{\left( n-1 \right)}^{2}} \right)+\left( n-1 \right) \right] \\
 & \Rightarrow S=\left( 2n-1 \right)\left[ 1+{{\left( n-1 \right)}^{2}}+n-1 \right] \\
 & \Rightarrow S=\left( 2n-1 \right)\left[ {{\left( n-1 \right)}^{2}}+n \right] \\
 & \Rightarrow S=\left( 2n-1 \right)\left[ {{n}^{2}}+1-2n+n \right] \\
 & \Rightarrow S=\left( 2n-1 \right)\left[ {{n}^{2}}-n+1 \right] \\
\end{align}\]

So, the correct answer is “$S=\left( 2n-1 \right)\left[ {{n}^{2}}-n+1 \right]$”.

Note: Now note that here when we are using the sum of AP our n changes. When we are calculating the sum of 1 + 2 + 3 + 4 + …. (n-1) terms. We substitute n – 1 in place of n. Similarly while finding the sum of 2n – 1 terms we substitute the value of n as 2n – 1.