Question

If ${S_1}$ , ${S_2}$ , ${S_3}$ are thirst n natural numbers, their squares, their cubes, respectively, then $\dfrac{{{S_3}(1 + 8{S_1})}}{{S_2^2}}$ is equal to
A.1
B.3
C.9
D.10

Answer 1

Hint : To solve this we need to know the formulas for sum of first ‘n’ natural number, sum of squares of first ‘n’ natural number and sum of cubes of first ‘n’ natural number. If we know these we can solve the given problem by substituting. We also use the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ while simplifying.

Complete step-by-step answer :
The sum of the first ‘n’ natural number $= \dfrac{{n(n + 1)}}{2}$ .
The sum of squares of the first ‘n’ natural number $= \dfrac{{n(n + 1)(2n + 1)}}{6}$ .
The sum of cubes of the first ‘n’ natural number $= \dfrac{{{n^2}{{(n + 1)}^2}}}{4}$ .
Now we have, $\dfrac{{{S_3}(1 + 8{S_1})}}{{S_2^2}}$
We solve the numerator term and the denominator term separately, then we substitute in the given problem we will get the required answer,
The numerator term $= {S_3}(1 + 8{S_1})$
Substituting, we get:
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times \left( {1 + 8\dfrac{{n(n + 1)}}{2}} \right)$
Cancelling 8 by 2 we get,
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times \left( {1 + 4n(n + 1)} \right)$
Expanding the brackets,
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times \left( {1 + 4{n^2} + 4n} \right)$
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times \left( {4{n^2} + {1^2} + 4n} \right)$
We know ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and $4{n^2} + {1^2} + 4n = {(2n + 1)^2}$ .
Above becomes,
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times {(2n + 1)^2}$
To further simplify multiply and divide by 36.
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{4} \times {(2n + 1)^2} \times \dfrac{{36}}{{36}}$
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{{4 \times 36}} \times {(2n + 1)^2} \times 36$
Cancelling 36 by 4, we get:
$= \dfrac{{{n^2}{{(n + 1)}^2}}}{{36}} \times {(2n + 1)^2} \times 9$
$= 9\left( {\dfrac{{{n^2}{{(n + 1)}^2}}}{{{6^2}}} \times {{(2n + 1)}^2}} \right)$
Since, square is present in each term in the bracket we have,
$= 9{\left( {\dfrac{{n(n + 1)}}{6} \times (2n + 1)} \right)^2}$
Now, we have the denominator term $= S_2^2$
$= {\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right)^2}$
Now $\dfrac{{{S_3}(1 + 8{S_1})}}{{S_2^2}}$ becomes
$= \dfrac{{9{{\left( {\dfrac{{n(n + 1)}}{6} \times (2n + 1)} \right)}^2}}}{{{{\left( {\dfrac{{n(n + 1)}}{6} \times (2n + 1)} \right)}^2}}}$
$= 9$
So, the correct answer is “Option C”.

Note : We simplified the above problem separately for avoiding the mistakes in cancellation or sign multiplication. We need to simplify the terms so that it will get canceled without simplification. We can prove the above formula of sum of first n natural number, sum of squares of n natural number, sum of cubic of n natural number by mathematical induction.