Question

Let ${S_n}$denote the sum of cubes of the first $n$ natural numbers and ${s_n}$ denote the sum of the first $n$ natural numbers. Then $\sum\nolimits_{r = 1}^n {\dfrac{{{S_r}}}{{{s_r}}}}$ is equal to
A.$\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{6}$
B.$\dfrac{{n\left( {n + 1} \right)}}{2}$
C.$\dfrac{{{n^2} + 3n + 2}}{6}$
D.None of these

Answer 1

The sum of first $n$ natural numbers has a well-known formula. The sum of cubes of first $n$ natural numbers can be observed by writing the first few values of the sum of the cubes of $n$ and can be easily correlated with the formula of the sum of first $n$ natural numbers.

Formula Used:
The formula for the sum of first $n$natural numbers is:
${s_n}$ = $\dfrac{{n\left( {n + 1} \right)}}{2}$ …(i)

Complete step-by-step answer:
As given in (i), the formula for the sum of first $n$ natural numbers is:
${s_n}$ = $\dfrac{{n\left( {n + 1} \right)}}{2}$
So, if we apply the formula for a few values, we have:
${s_5}$ = $\dfrac{{5\left( {5 + 1} \right)}}{2}$ = $\dfrac{{5 \times 6}}{2}$ = $15$ ${s_8}$ = $\dfrac{{8 \times \left( {8 + 1} \right)}}{2}$ = $\dfrac{{8 \times 9}}{2}$ = $36$
${s_{11}}$ = $\dfrac{{11 \times \left( {11 + 1} \right)}}{2}$ = $\dfrac{{11 \times 12}}{2}$ = $66$
Now, we do the sum of cubes of natural numbers of these many first $n$ natural numbers and see if we get some relation between the two:
$\Rightarrow$ ${S_5}$ = ${1^3} + {2^3} + {3^3} + {4^3} + {5^3}$ = $1 + 8 + 27 + 64 + 125$ = $225$
Similarly, for $n$ = $8$ it is:
$\Rightarrow$ ${S_8}$ = ${S_5} + {6^3} + {7^3} + {8^3}$ = $225 + 216 + 343 + 512$ = $1296$
And for $n$ = $11$ it is:
$\Rightarrow$ ${S_{11}}$ = ${S_8} + {9^3} + {10^3} + {11^3}$ = $1296 + 729 + 1000 + 1331$ = $4356$
Now, we look for the relation between ${S_n}$ and ${s_n}$, we can easily make out that the sum of the cubes is equal to the square of the sum of the numbers:
$\Rightarrow$ ${S_5}$ = $225$ = ${\left( {15} \right)^2}$ = ${\left( {{s_5}} \right)^2}$
and, ${S_8}$ = $1296$ = ${\left( {36} \right)^2}$ = ${\left( {{s_8}} \right)^2}$
and finally, ${S_{11}}$=$4356$=${\left( {66} \right)^2}$=${\left( {{s_{11}}} \right)^2}$
or, to put in a more general and a comprehensible formula, we can say that:
$\Rightarrow$ ${S_n}$ = ${\left( {{s_n}} \right)^2}$
Hence, we can also pen down the formula for ${S_n}$, which is:
$\Rightarrow$ ${S_n} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}$
So, the given question’s answer is as follows:
$\Rightarrow$ $\sum\nolimits_{r = 1}^n {\dfrac{{{S_r}}}{{{s_r}}}}$=$\dfrac{{{1^3} + {2^3} + {3^3} + ... + {{\left( {n - 1} \right)}^3} + {n^3}}}{{1 + 2 + 3 + ... + \left( {n - 1} \right) + n}}$
Using the above formulae into the expressions we have:
$\Rightarrow$ $\sum\nolimits_{r = 1}^n {\dfrac{{{S_r}}}{{{s_r}}}}$=$\dfrac{{{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}^2}}}{{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)}}$= $\dfrac{{n\left( {n + 1} \right)}}{2}$
Hence, the correct answer to the question is B$$$$\dfrac{{n\left( {n + 1} \right)}}{2}.

Note: So, we can see that solving these questions is very easy, one only needs to remember the formulae and then it’s going to be nothing but a simple division. If someone can not remember the formulae, then they can derive it as was shown above. And then, solve the question.