Question

If $C_0,C_1,C_2,\cdots ,C_n$ are the binomial coefficients and $S=2\times C_1+2^3\times C_3+2^5\times C_5+\cdots$, then which of the following is/are correct?

• A. When $n=50$, then $S=\frac{3^{50}-1}{2}$
• B. When $n=50$, then $S=\frac{3^{50}+1}{2}$
• C. When $n=101$, then $S=\frac{3^{101}+1}{2}$
• D. When $n=101$, then $S=\frac{3^{101}-1}{2}$

Answer 1

Answer: (A), (C)

The correct options are
A When $n=50$, then $S=\frac{3^{50}-1}{2}$
C When $n=101$, then $S=\frac{3^{101}+1}{2}$

$(1+x{)}^n=C_0+C_{1}x+C_{2}x+C_2{x}^2+C_3{x}^3+\cdots +C_n{x}^n$
$(1-x{)}^n=C_0-C_{1}x+C_2{x}^2-C_3{x}^3+\cdots +(-1{)}^n{C}_n{x}^n$
$\left[\right(1+x{)}^n-(1-x{)}^n]=2[C_{1}x+C_3{x}^3+C_5{x}^5+\cdots ]\frac{1}{2}\left[\right(1+x{)}^n-(1-x{)}^n]=C_{1}x+C_3{x}^3+C_5{x}^5+\cdots$
Putting $x=2$, we have $2\times C_1+2^3\times C_3+2^5\times C_5+\cdots =\frac{3^n-(-1{)}^n}{2}\Rightarrow S=\frac{3^n-(-1{)}^n}{2}$

When $n=50$,
$S=\frac{3^{50}-1}{2}$

When $n=101$
$S=\frac{3^{101}+1}{2}$