The cubes of natural numbers are grouped as $\left( {{1}^{3}} \right),\left( {{2}^{3}},{{3}^{3}} \right),\left( {{4}^{3}},{{5}^{3}},{{6}^{3}} \right),\cdots $, then the sum of the numbers in the nth group is
[a] $\dfrac{{{n}^{2}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+4 \right)}{12}$
[b] $\dfrac{{{n}^{3}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+3 \right)}{8}$
[c] $\dfrac{{{n}^{3}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+4 \right)}{8}$
[d] $\dfrac{{{n}^{2}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+4 \right)}{16}$
Answer
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Hint: Find the starting term of the nth group. Observe that the first group contains one element, the second group contains two elements and so on. Hence the number of elements used till we reach the nth group is $1+2+3+\cdots +n-1$. Use the fact that the sum of first n natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$. Hence find the starting element of the nth group and hence find the sum of the numbers in nth group. Use the fact that the sum of cubes of first n natural numbers is given by ${{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
Complete step-by-step answer:
The number of elements in the first group =1
The number of elements in the first and the second group combined = 1+2 = 3
The number of elements in the first, second and third group combined = 1+2+3 = 6
Observe that the second group starts with the cube of 1+1, the third group starts with the cube of 3+1 and so on.
Hence the nth group starts with the cube of $\left( 1+2+3+\cdots +n-1 \right)+1$
We know that the sum of first n-natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$
Hence, we have $1+2+3+\cdots +n-1=\dfrac{n\left( n-1 \right)}{2}$
Hence, we have nth group starts with the cube of $\dfrac{n\left( n-1 \right)}{2}+1$
Hence, the nth group is
$\left( {{\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}^{3}},{{\left( \dfrac{n\left( n-1 \right)}{2}+2 \right)}^{3}},\cdots ,{{\left( \dfrac{n\left( n-1 \right)}{2}+n \right)}^{3}} \right)$
Hence, we have sum of elements in the nth group is given by
${{S}_{n}}={{\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}^{3}}+{{\left( \dfrac{n\left( n-1 \right)}{2}+2 \right)}^{3}}+\cdots +{{\left( \dfrac{n\left( n-1 \right)}{2}+n \right)}^{3}}$
Adding and subtracting ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdots +{{\left( \dfrac{n\left( n-1 \right)}{2} \right)}^{3}}$, we get
${{S}_{n}}=\sum\limits_{r=1}^{\dfrac{n\left( n-1 \right)}{2}+n}{{{r}^{3}}}-\sum\limits_{r=1}^{\dfrac{n\left( n-1 \right)}{2}}{{{r}^{3}}}$
We have $\dfrac{n\left( n-1 \right)}{2}+n=\dfrac{n}{2}\left( n-1+2 \right)=\dfrac{n\left( n+1 \right)}{2}$
Hence, we have
${{S}_{n}}=\sum\limits_{r=1}^{\dfrac{n\left( n+1 \right)}{2}}{{{r}^{3}}}-\sum\limits_{1}^{\dfrac{n\left( n-1 \right)}{2}}{{{r}^{3}}}$
We know that $\sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
Hence, we have
${{S}_{n}}={{\left[ \dfrac{\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)}{2} \right]}^{2}}-{{\left[ \dfrac{\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}{2} \right]}^{2}}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Hence, we have
$\begin{align}
& {{S}_{n}}=\dfrac{1}{4}\left[ \left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]\left[ \left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right] \\
& \Rightarrow {{S}_{n}}=\dfrac{{{S}_{1}}{{S}_{2}}}{4} \\
\end{align}$
In the first sum, we have
${{S}_{1}}=\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)$
Taking $\dfrac{n}{2}$ common, we get
${{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]$
Hence, we have
$\begin{align}
& {{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2} \right)+n+1-\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2} \right)-n+1 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( {{\left( n+1 \right)}^{2}}-{{\left( n-1 \right)}^{2}} \right)+2 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( 4n \right)+2 \right] \\
& =n\left( {{n}^{2}}+1 \right) \\
\end{align}$
In the second sum, we have
${{S}_{2}}=\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)$
Taking $\dfrac{n}{2}$ common, we get
${{S}_{2}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]$
Hence, we have
$\begin{align}
& {{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2} \right)+n+1+\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2} \right)+n-1 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( {{\left( n+1 \right)}^{2}}+{{\left( n-1 \right)}^{2}} \right)+2n \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( 2{{n}^{2}}+2 \right)+2n \right] \\
& =\dfrac{n\left( {{n}^{3}}+3n \right)}{2} \\
& =\dfrac{{{n}^{2}}\left( {{n}^{2}}+3 \right)}{2} \\
\end{align}$
Hence, we have
$\begin{align}
& {{S}_{n}}=\dfrac{n\left( {{n}^{2}}+1 \right)\left( {{n}^{2}} \right)\left( {{n}^{2}}+3 \right)}{8} \\
& =\dfrac{{{n}^{3}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+3 \right)}{8} \\
\end{align}$
Hence option [b] is correct.
Note: Verification:
Check if it is correct for the first three groups
${{S}_{1}}=1$
Put n = 1 in the expression of ${{S}_{n}}$ , we get
${{S}_{1}}=\dfrac{1}{8}\left( 1+1 \right)\left( 1+3 \right)=1$
We have
${{S}_{2}}={{2}^{3}}+{{3}^{3}}=35$
Put n =2 in the expression of ${{S}_{n}}$, we get
${{S}_{2}}=\dfrac{8}{8}\left( 4+1 \right)\left( 4+3 \right)=35$
We have ${{S}_{3}}=\left( {{4}^{3}}+{{5}^{3}}+{{6}^{3}} \right)=64+125+216=405$
Put n =3 in the expression of ${{S}_{n}}$, we get
${{S}_{3}}=\dfrac{27}{8}\left( 9+1 \right)\left( 9+3 \right)=405$
Hence our answer is verified to be correct.
Complete step-by-step answer:
The number of elements in the first group =1
The number of elements in the first and the second group combined = 1+2 = 3
The number of elements in the first, second and third group combined = 1+2+3 = 6
Observe that the second group starts with the cube of 1+1, the third group starts with the cube of 3+1 and so on.
Hence the nth group starts with the cube of $\left( 1+2+3+\cdots +n-1 \right)+1$
We know that the sum of first n-natural numbers is given by $\dfrac{n\left( n+1 \right)}{2}$
Hence, we have $1+2+3+\cdots +n-1=\dfrac{n\left( n-1 \right)}{2}$
Hence, we have nth group starts with the cube of $\dfrac{n\left( n-1 \right)}{2}+1$
Hence, the nth group is
$\left( {{\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}^{3}},{{\left( \dfrac{n\left( n-1 \right)}{2}+2 \right)}^{3}},\cdots ,{{\left( \dfrac{n\left( n-1 \right)}{2}+n \right)}^{3}} \right)$
Hence, we have sum of elements in the nth group is given by
${{S}_{n}}={{\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}^{3}}+{{\left( \dfrac{n\left( n-1 \right)}{2}+2 \right)}^{3}}+\cdots +{{\left( \dfrac{n\left( n-1 \right)}{2}+n \right)}^{3}}$
Adding and subtracting ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\cdots +{{\left( \dfrac{n\left( n-1 \right)}{2} \right)}^{3}}$, we get
${{S}_{n}}=\sum\limits_{r=1}^{\dfrac{n\left( n-1 \right)}{2}+n}{{{r}^{3}}}-\sum\limits_{r=1}^{\dfrac{n\left( n-1 \right)}{2}}{{{r}^{3}}}$
We have $\dfrac{n\left( n-1 \right)}{2}+n=\dfrac{n}{2}\left( n-1+2 \right)=\dfrac{n\left( n+1 \right)}{2}$
Hence, we have
${{S}_{n}}=\sum\limits_{r=1}^{\dfrac{n\left( n+1 \right)}{2}}{{{r}^{3}}}-\sum\limits_{1}^{\dfrac{n\left( n-1 \right)}{2}}{{{r}^{3}}}$
We know that $\sum\limits_{r=1}^{n}{{{r}^{3}}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$
Hence, we have
${{S}_{n}}={{\left[ \dfrac{\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)}{2} \right]}^{2}}-{{\left[ \dfrac{\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)}{2} \right]}^{2}}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Hence, we have
$\begin{align}
& {{S}_{n}}=\dfrac{1}{4}\left[ \left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]\left[ \left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right] \\
& \Rightarrow {{S}_{n}}=\dfrac{{{S}_{1}}{{S}_{2}}}{4} \\
\end{align}$
In the first sum, we have
${{S}_{1}}=\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)$
Taking $\dfrac{n}{2}$ common, we get
${{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)-\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]$
Hence, we have
$\begin{align}
& {{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2} \right)+n+1-\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2} \right)-n+1 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( {{\left( n+1 \right)}^{2}}-{{\left( n-1 \right)}^{2}} \right)+2 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( 4n \right)+2 \right] \\
& =n\left( {{n}^{2}}+1 \right) \\
\end{align}$
In the second sum, we have
${{S}_{2}}=\left( \dfrac{n\left( n+1 \right)}{2} \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( \dfrac{n\left( n-1 \right)}{2} \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right)$
Taking $\dfrac{n}{2}$ common, we get
${{S}_{2}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2}+1 \right)+\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2}+1 \right) \right]$
Hence, we have
$\begin{align}
& {{S}_{1}}=\dfrac{n}{2}\left[ \left( n+1 \right)\left( \dfrac{n\left( n+1 \right)}{2} \right)+n+1+\left( n-1 \right)\left( \dfrac{n\left( n-1 \right)}{2} \right)+n-1 \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( {{\left( n+1 \right)}^{2}}+{{\left( n-1 \right)}^{2}} \right)+2n \right] \\
& =\dfrac{n}{2}\left[ \dfrac{n}{2}\left( 2{{n}^{2}}+2 \right)+2n \right] \\
& =\dfrac{n\left( {{n}^{3}}+3n \right)}{2} \\
& =\dfrac{{{n}^{2}}\left( {{n}^{2}}+3 \right)}{2} \\
\end{align}$
Hence, we have
$\begin{align}
& {{S}_{n}}=\dfrac{n\left( {{n}^{2}}+1 \right)\left( {{n}^{2}} \right)\left( {{n}^{2}}+3 \right)}{8} \\
& =\dfrac{{{n}^{3}}\left( {{n}^{2}}+1 \right)\left( {{n}^{2}}+3 \right)}{8} \\
\end{align}$
Hence option [b] is correct.
Note: Verification:
Check if it is correct for the first three groups
${{S}_{1}}=1$
Put n = 1 in the expression of ${{S}_{n}}$ , we get
${{S}_{1}}=\dfrac{1}{8}\left( 1+1 \right)\left( 1+3 \right)=1$
We have
${{S}_{2}}={{2}^{3}}+{{3}^{3}}=35$
Put n =2 in the expression of ${{S}_{n}}$, we get
${{S}_{2}}=\dfrac{8}{8}\left( 4+1 \right)\left( 4+3 \right)=35$
We have ${{S}_{3}}=\left( {{4}^{3}}+{{5}^{3}}+{{6}^{3}} \right)=64+125+216=405$
Put n =3 in the expression of ${{S}_{n}}$, we get
${{S}_{3}}=\dfrac{27}{8}\left( 9+1 \right)\left( 9+3 \right)=405$
Hence our answer is verified to be correct.
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