The value of _r=1^15 r^2(^15C_r^15C_r-1) is equal to:

Question

The value of $\displaystyle\sum_{r=1}^{15} r^2\left(\frac{^{15}C_r}{{}^{15}C_{r-1}}\right)$ is equal to:

Accepted Answer

We know,

$${{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}$$

$$ \therefore $$    $${{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}$$

$$ \therefore $$     $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)$$

$=$ $$\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)$$

$$ = \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)} $$

$$ = 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} } $$

$$ = 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}$$

$=$  120 $$ \times $$ 16 - 40 $$ \times $$ 31

$=$  680

Solutions · 1