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., $(a+b)^n$ = $\sum\limits_{r=0}^{n} \,\binom{n}{r} a^{n-r} b^r$ Properties of binomial coefficients In the binomial expansion of (1 + x) n the coefficients $\,\binom{n}{0} , \,\binom{n}{1},\,\binom{n}{2}$ , ... , $\,\binom{n}{n}$ are denoted by $C_0,C_1,C_2,\ldots,C_n$ respectively $C_0+C_1+C_2+\cdots+C_n$ = $2^n$ We have $(1+x)^{50}$ $(1+x)^{50}$ = $\sum\limits_{r=0}^{50} \,\binom{50}{r} x^r$ (Since by binomial expansion) ∴ sum of the coefficients of odd power of x = $\,\binom{50}{1} + \,\binom{50}{3}$ + ... + $\,\binom{50}{49}$ = (Since $C_0+C_2+C_4$ + ... = $C_1+C_3+C_5$ + ... = 2n - 1) = $\frac{1}{2} \times 2^{50}$ (since $C_0 + C_1 + C_2$ + ... + $C_{50}$ = $2^n$) = $2^{50} \times 2^{-1}$ = $2^{50-1}$ = $2^{49}$ Hence, In the expansion of $(1+x)^{50}$ the sum of the cofficients of the odd powers of x is $2^{49}$ Another method : We know that To find the sum of coefficients of odd powers of x, we find the values of the expression by replacing the variables by 1 and - 1 respectively and the subtracting the two results Let f (x) = $(1+x)^{50}$ Replacing the variables by 1 and - 1 f (1) = $(1+1)^{50}$ = $2^{50}$ f (2) = $(1-1)^{50}$ = 0 By subtracting the two results we get the odd power of x $\frac{f(1)-f(-1)}{2}$ = $\frac{2^{50}-0}{2}$ = $\frac{2^{50}}{2}$ = $2^{50}\times 2^{-1}$ = $2^{49}$ Hence In the expansion of $(1+x)^{50}$ the sum of the coefficients of the odd powers of x is $2^{49}$.