JEE Main 2023 (Online) 25th January Morning ShiftQuestion 1
If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\sum_{r=1}^{10}r^{3}\left(\frac{a_r}{a_{r-1}}\right)^{2}$ is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. By symmetry $a_r={}^{10}C_r$; the ratio of consecutive coefficients turns the sum into a polynomial sum. Formulas used. (1) $\dfrac{{}^{n}C_r}{{}^{n}C_{r-1}}=\dfrac{n-r+1}{r}$(2) $\sum_{r=1}^{10}r=55$, $\sum r^2=385$, $\sum r^3=3025$Step 1 - $a_r$. The coefficient of $x^{10-r}$ is ${}^{10}C_{10-r}={}^{10}C_r$. By (1), $\dfrac{a_r}{a_{r-1}}=\dfrac{11-r}{r}$. Step 2 - Simplify the term. $$r^3\left(\frac{11-r}{r}\right)^2=r(11-r)^2=121r-22r^2+r^3.$$ Step 3 - Sum. By (2), $$121\cdot55-22\cdot385+3025=6655-8470+3025=1210.$$ Answer: (B)