Question

If the sum of the series ${}^{12}{C}_5+7{.}^{12}{C}_6+21{.}^{12}{C}_7+35{.}^{12}{C}_8+35{.}^{12}{C}_9+21{.}^{12}{C}_{10}+{7.}^{12}{C}_{11}{+}^{12}{C}_{12}$ is equal to ${}^{2n+1}{C}_{12}$ then $n$ is

Answer 1

Series is ${}^7{C}_7{.}^{12}{C}_5{+}^7{C}_6{.}^{12}{C}_6{+}^7{C}_5{.}^{12}{C}_7{+}^7{C}_4{.}^{12}{C}_8{+}^7{C}_3{.}^{12}{C}_9{+}^7{C}_2{.}^{12}{C}_{10}{+}^7{C}_1{.}^{12}{C}_{11}{+}^7{C}_0{.}^{12}{C}_{12}$
$(1+x{)}^7{=}^7{C}_0{+}^7{C}_{1}x{+}^7{C}_2{x}^2+------{+}^7{C}_7{x}^7$
$(1+x{)}^{12}{=}^{12}{C}_0{+}^{12}{C}_{1}x{+}^{12}{C}_2{x}^2{+}^{12}{C}_5{x}^5-{+}^{12}{C}_{12}{x}^{12}$
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required series is equal to coeff. of $x^{12}$ in multiplication of above two series.

So coeff. of $x^{12}$ in $(1+x{)}^{19}\text{is}{}^{19}{C}_{12}.$