Wow... it's amazing

We know that :

${S}_{1}=\frac{n(n+1)}{2}$

${S}_{2}=\frac{n(n+1)(2n+1)}{6}$

${S}_{3}=\frac{{n}^{2}{(n+1)}^{2}}{4}$

Just plug in the values to get :

$=\frac { \frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } (1+8\frac { n(n+1) }{ 2 } ) }{ { (\frac { n(n+1)(2n+1) }{ 6 } ) }^{ 2 } }$

$=9$

Take n=1, s1= s2= s3= 1. so the value of expression= 9

Use the direct formulas for the sum of first n numbers, sum of squares of first n number and sum of cubes of first n numbers. This will give answer straightaway.

Now, can this be solved if we dont remember the formulas? Yes we can. Becasue this is a multiple choice qn, trial and error is an appropriate method.

Now, let the numbers be the first 3 natural numbers (1,2,3)

Then we get,

S1= 1+2+3 = 6.

S2 = 1+4+9 = 14,

S3 = 1+8+27 = 36.

Now putting these values in

$\frac{S_3(1+8S_1)}{(S_2)^2}=9$

Now to be sure of the answer, try with a different pattern. Take the first 5 natural numbers (1,2,3,4 and 5)

Then we get,

S1= 1+2+3+4+5 = 15.

S2 = 1+4+9+16+25 = 55,

S3 = 1+8+27+64+125 = 225.

Now putting these values in

$\frac{S_3(1+8S_1)}{(S_2)^2}=9$

So, the answer is 9

s3=(s1)^2; s2=s1(2n+1)/3

Plugging in the values we get the answer=9

Straightforward using a computer symbolic algebra software, in this case the Python sympy library:

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>>> from sympy import *
>>> var('n')
n
>>> S1,S2,S3=symbols('S1 S2 S3')
>>> S1=n*(n+1)/2
>>> S3=S1**2
>>> S2=n*(n+1)*(2*n+1)/6
>>> (S3*(1+8*S1)).expand()
n**6 + 3*n**5 + 13*n**4/4 + 3*n**3/2 + n**2/4
>>> (S2**2).expand()
n**6/9 + n**5/3 + 13*n**4/36 + n**3/6 + n**2/36
>>> div(n**6 + 3*n**5 + 13*n**4/4 + 3*n**3/2 + n**2/4,n**6/9 + n**5/3 + 13*n**4/36 + n**3/6 + n**2/36)
(9, 0)
>>>