Binomial Theorem
Binomial Theorem
star_batch_jee_advanced_2025
Grade None
Question:
If $a_n = \sum_{r=0}^{n} \frac{1}{C_r}$, then $\sum_{r=0}^{n} \frac{r^2}{C_r} = P(n)a_{n+2} + Q(n)a_{n+1} + a_n + R(n)$, where $P(n), Q(n), R(n)$ are the polynomial functions of $n$, then $P(5) =$
Step-by-Step Solution
Key Concept: Use the recurrence relation $\frac{r^2}{\binom{n}{r}} = \frac{r^2 \cdot r!(n-r)!}{n!}$ combined with the identity $r^2 \binom{n}{r} = n(n-1)\binom{n-2}{r-2} + n\binom{n-1}{r-1}$ to express $\sum_{r=0}^{n} \frac{r^2}{C_r}$ in terms of shifts of $a_n$, relating it to $a_{n+2}$, $a_{n+1}$, and $a_n$.
Since the roots of $a^2 + b^2 - 1 0$.
Correct Answer: I need to find the relationship for $\sum_{r=0}^{n} \frac{r^2}{C_r}$ and determine $P(5)$.
Given: $a_n = \sum_{r=0}^{n} \frac{1}{C_r}$ where $C_r = \binom{