A) 4
B) 8
C) 16
D) 32
Correct Answer: A
Solution :
We have \[^{2n+1}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}=255\] ? (i) Also the sum of binomial coefficients \[^{2n+1}{{C}_{0}}{{+}^{2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}\] \[{{+}^{2n+1}}{{C}_{n+1}}+{{+}^{2n+1}}{{C}_{n+2}}+...{{+}^{2n+1}}{{C}_{2n+1}}\] \[={{(1+1)}^{2n+1}}={{2}^{2n+1}}\] \[\Rightarrow \]\[^{2n+1}{{C}_{0}}+2{{\{}^{2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}\}\] \[{{+}^{2n+1}}{{C}_{2n+1}}={{2}^{2n+1}}\] \[\Rightarrow \]\[1+2(255)+1={{2}^{2n+1}}\] \[\Rightarrow \] \[1+255={{2}^{2n}}\] \[\Rightarrow \] \[{{2}^{2n}}={{2}^{8}}\] \[\Rightarrow \] \[n=4\]You need to login to perform this action.
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