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

Hint: 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.