JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
If $p_1$ and $p_2$, respectively, are the second largest and the fourth largest prime numbers in the factorization of the coefficient of $x^{30}$ in the expansion of $\left(x^2+2+\frac1{x^2}\right)^{-5}(1+x^2)^{40}$, then $p_1+p_2$ is equal to ________.
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$34$
Solution
Concept. $x^2+2+\frac1{x^2}$ is a perfect square, so the whole expression is a single binomial. Formulas used. (1) $x^2+2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2=\dfrac{(1+x^2)^2}{x^2}$Step 1 - Simplify. By (1), $$\left(x^2+2+\frac1{x^2}\right)^{-5}(1+x^2)^{40}=\frac{x^{10}}{(1+x^2)^{10}}(1+x^2)^{40}=x^{10}(1+x^2)^{30}.$$ Step 2 - The coefficient. We need $x^{20}$ in $(1+x^2)^{30}$, i.e. $(x^2)^{10}$: ${}^{30}C_{10}$. Step 3 - Factorise. ${}^{30}C_{10}=\dfrac{21\cdot22\cdots30}{10!}$. Count each prime in the numerator and subtract $10!=2^8\cdot3^4\cdot5^2\cdot7$: $${}^{30}C_{10}=3^2\cdot5\cdot7\cdot11\cdot13\cdot23\cdot29=30045015.$$ Step 4 - Read off. The primes, largest first, are $29,23,13,11,\dots$ So $p_1=23$, $p_2=11$, and $p_1+p_2=34$. Answer: 34