Question

The sum \(1 + \frac{{{1^3} + {2^3}}}{{1 + 2}} + \frac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + \ldots + \frac{{{1^3} + {2^3} + {3^3} + \ldots + {{15}^3}}}{{1 + 2 + 3 + \ldots + 15}} - \frac{1}{2}\left( {1 + 2 + 3 + \ldots + 15} \right)\) is equal to:
1. 620
2. 1240
3. 1860
4. 660

Answer 1

Correct Answer - Option 1 : 620

Let \({\rm{S}} = 1 + \frac{{{1^3} + {2^3}}}{{1 + 2}} + \frac{{{1^3} + {2^3} + {3^3}}}{{1 + 2 + 3}} + \ldots 15{\rm{\;terms}}\) 

\({T_n} = \frac{{{1^3} + {2^3} + \ldots {n^3}}}{{1 + 2 + \ldots n}} = \frac{{{{\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)}^2}}}{{\frac{{n\left( {n + 1} \right)}}{2}}} = \frac{{n\left( {n + 1} \right)}}{2}\) 

Now, \(S = \frac{1}{2}\left( {\mathop \sum \limits_{n = 1}^{15} {n^2} + \mathop \sum \limits_{n = 1}^{15} n} \right)\) 

\( = \frac{1}{2} \times \frac{{15 \times 16 \times 31}}{6} + \frac{1}{2} \times \frac{{15 \times 16}}{2}\) 

= 620 + 60 = 680

∴ Required sum \( = 680 - \frac{1}{2}\frac{{15 \times 16}}{2} = 680 - 60 = 620\)