Question

The coefficient of $t^{50}$ in ${(1+t^2)}^{25}\left({1+t}^{25}\right)(1+t^{40})(1+t^{45})(1+t^{47})$ is

• A. $1+{{}^{25}C}_5$
• B. $1+{{}^{25}C}_5+{{}^{25}C}_7$
• C. $1+{{}^{25}C}_7$
• D. None of these

Answer 1

The correct option is A $1+{{}^{25}C}_5$

$(1+t^2{)}^{25}(1+t^{25}+t^{40}+t^{45}+t^{47})$ ... (upon ignoring the terms with power more than $50$)
$=(1+{}^{25}{C}_1{t}^2+{}^{25}{C}_2{t}^4+{}^{25}{C}_3{t}^6+...{}^{25}{C}_{25}{t}^{50})(1+t^{25}+t^{40}+t^{45}+t^{47})$
Now $t^{50}$ can be achieved in the following ways from the above expression $(t^2{)}^{25}.1$ or $(t^2{)}^5.t^{40}$
Hence the coefficient of $t^{50}$ will be
$=$ Sum of the coefficients of the above individual terms.
$={}^{25}{C}_{25}+{}^{25}{C}_5$
$=1+{}^{25}{C}_5$.