Binomial Theorem Question 5
Question 5 - 25 January - Shift 1
If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\sum _{r=1}^{10} r^{3}(\frac{a_r}{a _{r-1}})^{2}$ is equal to
(1) 4895
(2) 1210
(3) 5445
(4) 3025
Show Answer
Answer: (2)
Solution:
Formula: Properties of Binomial Theorem for Positive Integer
$a_r={ }^{10} C _{10-r}={ }^{10} C_r$
$\Rightarrow \sum _{r=1}^{10} r^{3}(\frac{{ }^{10} C_r}{{ }^{10} C _{r-1}})^{2}=\sum _{r=1}^{10} r^{3}(\frac{11-r}{r})^{2}=\sum _{r=1}^{10} r(11-r)^{2}$
$=\sum _{r=1}^{10}(121 r+r^{3}-22 r^{2})=1210$
