Question

If $\sum _{r=1}^{10}r(r-1{)}^{10}{C}_r=k{.2}^9$, then $k$ is equal to

• A. $10$
• B. $45$
• C. $90$
• D. $100$

Answer 1

The correct option is B $45$

${\sum }_{r=1}^{10}r(r-1{)}^{10}{C}_r$

${}^{10}{C}_r={\frac{10}{r}}^9{C}_{r-1}$

$=\frac{10}{r}\times \frac{9}{r-1}{}^8{C}_{r-2}$

$=\sum r(r-1)\times \frac{10}{r}\times {\frac{9}{r-1}}^8{C}_{r-2}$

$=90{\sum }_{r=2}^{10}{C}_{r-2}$

$=45\left[2^9\right]=K\cdot 2^9$

$K=45$