JEE Main 2020 (Online) 5th September Evening ShiftQuestion 1
The coefficient of $x^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{6}$ in powers of $x$, is $\_\_\_\_\_\_\_\_$ .
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$120$
Solution
Concept. $1+x+x^2+x^3=(1+x)(1+x^2)$, so the power splits into two simple binomials. Formulas used. (1) $1+x+x^2+x^3=(1+x)(1+x^2)$Step 1 - Split. By (1), the expression is $(1+x)^6(1+x^2)^6$. Step 2 - Ways to get $x^4$. Take $x^{2j}$ from $(1+x^2)^6$ and $x^{4-2j}$ from $(1+x)^6$: $$j=0:\ {}^{6}C_4=15,\qquad j=1:\ {}^{6}C_1\cdot{}^{6}C_2=90,\qquad j=2:\ {}^{6}C_2=15.$$ Step 3 - Add. $15+90+15=120$. Answer: 120