Question

The value of $\sum_{r = 1}^{10}{r \cdot \frac{nC_{r}}{nC_{r - 1}}}$ is equal to

• A. 5(2n – 9)
• B. 10n
• C. 9(n – 4)
• D. None

Answer 1

Answer: (A)

5(2n – 9)

Answer 2

$\sum_{r = 1}^{10}{r.}\frac{nC_{r}}{nC_{r–1}}$, $\sum_{r = 1}^{10}{r.}\frac{n–r + 1}{r}$, (n + 1) $\sum_{r = 1}^{10}1–\sum_{r = 1}^{10}r$

(n + 1) (10) – $\frac{10(10 + 1)}{2}$ = 10n + 10 – 55

= 10n – 45 = 5 (2n – 9)