JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
Let $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$. Then $\frac{a_7}{a_{13}}$ is equal to ________.
Binomial Theorem · Multinomial Expansion
Answer$8$
Solution
Concept. Replacing $x$ by $\frac2y$ turns $2x^2+3x+4$ into a multiple of itself, which links $a_r$ with $a_{20-r}$. Formulas used. (1) $2\left(\frac2y\right)^2+3\cdot\frac2y+4=\dfrac{2\left(2y^2+3y+4\right)}{y^2}$Step 1 - Substitute. Let $P(x)=(2x^2+3x+4)^{10}=\sum a_rx^r$. By (1), $$P\left(\frac2y\right)=\frac{2^{10}\,P(y)}{y^{20}}.$$ Step 2 - Compare coefficients. The left side is $\sum a_r\,2^r\,y^{-r}$; the right side is $2^{10}\sum a_s\,y^{s-20}$. Matching $y^{-r}$ (so $s=20-r$): $$a_r\,2^r=2^{10}\,a_{20-r}.$$ Step 3 - Put $r=7$. $a_7\cdot2^7=2^{10}\,a_{13}$, so $\dfrac{a_7}{a_{13}}=2^3=8$. Answer: 8