Question

Sum up to 16 terms of the series $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+\cdots \cdots $ is
\[\begin{align}
  & A.450 \\
 & B.456 \\
 & C.446 \\
 & \text{D}.\text{None of these} \\
\end{align}\]

Answer 1

Hint: For solving the sum of 16 term series $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+\cdots \cdots $ we will first find ${{n}^{th}}$ term of the series denoted by ${{T}_{n}}$. Then we will find a sum of n terms of series denoted by ${{S}_{n}}$ using $\sum{{{T}_{n}}}$. We will use following formula for solving this sum,
(1) ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdots \cdots +{{n}^{3}}=\sum{{{n}^{3}}}\text{ and }\sum{{{n}^{3}}}$ is equal to ${{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$.
(2) For an AP, $a,a+d,a+2d,\ldots \ldots a+\left( n-1 \right)d$ sum of n terms is given by $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
(3) $\sum \left( {{a}_{n}}+{{b}_{n}} \right)=\sum {{a}_{n}}+\sum {{b}_{n}}$.
(4) Sum ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\cdots \cdots +{{n}^{2}}=\sum{{{n}^{2}}}$ is equal to $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.
(5) Sum $1+2+3+\ldots \ldots n=\sum n$ is equal to $\dfrac{n\left( n+1 \right)}{2}$.
(6) Sum $1+1+1+\ldots \ldots n=\sum 1$ is equal to n.
Then we will put the value of n as 16 to get our final answer.

Complete step-by-step solution
Here, we are given the series as $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+\cdots \cdots $.
Let us first find the ${{n}^{th}}$ term of the series, for this, we will find ${{n}^{th}}$ term of numerator and denominator separately.
For numerator, ${{n}^{th}}$ term will be ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdots \cdots +{{n}^{3}}=\sum{{{n}^{3}}}$ which is equal to $\sum{{{n}^{3}}}$.
As we know $\sum{{{n}^{3}}}={{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$.
So, number of ${{n}^{th}}$ term is ${{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}$.
For denominators, since all odd terms are adding, so ${{n}^{th}}$ term will be $1+3+5+\ldots \ldots +\left( 2n-1 \right)$.
As we can see, this is an AP with the first term (a) as 1 and a common difference d(3-1=5-3=2) as 2. So let us find the sum of n terms of this AP. Since the sum of n terms of an AP is given by ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
So, we get,
${{S}_{n}}=\dfrac{n}{2}\left( 2+\left( n-1 \right)2 \right)\Rightarrow \dfrac{n}{2}\left( 2n \right)={{n}^{2}}$.
Hence the denominator of ${{n}^{th}}$ term is ${{n}^{2}}$.
So our ${{n}^{th}}$ term of series denoted by ${{T}_{n}}$ is $\dfrac{{{\left( \dfrac{n\left( n+1 \right)}{2} \right)}^{2}}}{{{n}^{2}}}\Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4{{n}^{2}}}=\dfrac{{{\left( n+1 \right)}^{2}}}{4}$.
Since ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ so,
\[\begin{align}
  & {{T}_{n}}=\dfrac{{{n}^{2}}+1+2n}{4}=\dfrac{{{n}^{2}}}{4}+\dfrac{1}{4}+\dfrac{2n}{4} \\
 & \Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{4}+\dfrac{n}{2}+\dfrac{1}{4} \\
\end{align}\]
Now we need to find sum of n terms of a series where ${{n}^{th}}$ term is ${{T}_{n}}=\dfrac{{{n}^{2}}}{4}+\dfrac{n}{2}+\dfrac{1}{4}$.
So we just need to find $\sum{{{T}_{n}}}$.
Let us denote the sum of n terms by ${{S}_{n}}$. Hence,
${{S}_{n}}=\sum{{{T}_{n}}}\Rightarrow {{S}_{n}}=\sum{\left( \dfrac{{{n}^{2}}}{4}+\dfrac{n}{2}+\dfrac{1}{4} \right)}$.
As we know, $\sum \left( {{a}_{n}}+{{b}_{n}} \right)=\sum {{a}_{n}}+\sum {{b}_{n}}$ so we get:
\[\begin{align}
  & {{S}_{n}}=\sum{\dfrac{{{n}^{2}}}{4}}+\sum{\dfrac{n}{2}}+\sum{\dfrac{1}{4}} \\
 & \Rightarrow {{S}_{n}}=\dfrac{1}{4}\sum{{{n}^{2}}}+\dfrac{1}{2}\sum{n}+\dfrac{1}{4}\sum{1} \\
\end{align}\]
Now we know that, sum of squares of n terms is given by $\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.
Also, the sum of n terms is given by \[\sum{n}=\dfrac{n\left( n+1 \right)}{2}\].
And the sum of $1+1+1+\ldots \ldots n$ is given by $\sum 1=n$.
So putting all these values we get:
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{1}{4}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)+\dfrac{1}{2}\left( \dfrac{n\left( n+1 \right)}{2} \right)+\dfrac{1}{4}\left( n \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{24}+\dfrac{n\left( n+1 \right)}{4}+\dfrac{n}{4} \\
\end{align}\]
Now, we need to find sum of first 16 terms, so putting n = 16, we get:
\[\begin{align}
  & \Rightarrow {{S}_{16}}=\dfrac{16\left( 16+1 \right)\left( 2\left( 16 \right)+1 \right)}{24}+\dfrac{16\left( 16+1 \right)}{4}+\dfrac{16}{4} \\
 & \Rightarrow {{S}_{16}}=\dfrac{16\times 17\times 33}{24}+\dfrac{16\times 17}{4}+4 \\
\end{align}\]
Simplifying we get:
\[\begin{align}
  & \Rightarrow {{S}_{16}}=374+68+4 \\
 & \Rightarrow {{S}_{16}}=446 \\
\end{align}\]
Hence sum of 16 terms of $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+\cdots \cdots $ is 446.
Hence option C is the correct answer.

Note: Students should know formulas of summation of some of the series such as $\sum n,\sum {{n}^{2}},\sum {{n}^{3}}$. Here we have not taken a sum of 16 sums directly as it would cause high calculations. Always try to make a general solution for the sum of a series and then put the value as 16. Here we have taken ${{S}_{n}}=\sum {{T}_{n}}$ which means we will take sum of all ${{T}_{n}}$ starting from 1 to n. Since ${{T}_{n}}=\dfrac{{{n}^{2}}}{4}+\dfrac{n}{2}+\dfrac{1}{4}$ which is equivalent to $\dfrac{{{1}^{3}}}{1}+\dfrac{{{1}^{3}}+{{2}^{3}}}{1+3}+\dfrac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+\cdots \cdots \dfrac{{{n}^{3}}}{\left( 2n-1 \right)}$.