UPSEE Solved Model Paper 6

Here we can find the coefficient of $x^{53}$ in the given expression We know that Binomial theorem for positive integral index If a , b are any two real numbers and n any natural number then $(a+b)^n$ = $\,\binom{n}{0} a^n b^0$ + $\,\binom{n}{1} a^{n-1} b$ + $\,\binom{n}{2} a^{n-2} b^2$ + ... + $\,\binom{n}{r} a^{n-r} b^r$ + ... + $\,\binom{n}{n} x^0 b^n$ , where $\,\binom{n}{r}$ = $\frac{n!}{(n-r)! r!}$ Here $\,\binom{n}{0},\,\binom{n}{1},\,\binom{n}{2}$ ... $\,\binom{n}{n}$ are called binomial coefficients i.e., $(a+b)^n$ = $\sum\limits_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ Given $\sum\limits_{m=0}^{100} 100 C_m (x-3)^{100-m} \cdot 2^m$ On expanding the summation , we have , = $100C_0 (x-3)^{100}+100C_1 (x-3)^{99} \cdot 2$ + $100C_2 (x-3)^{98} \cdot 2^2$ + ... + $100C_{100} \cdot 2^{100}$ = $[(x-3)+2]^{100}$ = $(x-1)^{100}$ = $(1-x)^{100}$ (Since, the power is even, by binomial expansion) Therefore, coefficient of $x^{53}$ = $(-1)^{53} \cdot \binom{100}{53}$ = $-\binom{100}{53}$ Hence, The coefficient of $x^{53}$ in the expression of $\sum\limits_{m=0}^{100} 100C_m (x-3)^{100-m} \cdot 2^m$ = $-\binom{100}{53}$ another method : we have $\sum\limits_{m=0}^{100} 100 C_m (x-3)^{100-m} \cdot 2^m$ = $[(x-3)+2]^{100}$ = $(1-x)^{100}$ Therefore, coefficient of $x^{53}$ = $\binom{100}{53} (-1)^{53}$ = $-\binom{100}{53}$ $\sum\limits_{m=0}^{100} 100 C_m (x-3)^{100-m} \cdot 2^m$ = $-\binom{100}{53}$