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Choose the correct option regarding the following statements
1. \(C_0+C_2+C_4+\ldots+C_n=2^{n-1}\), if \(n\) is even
2. \(C_1+C_3+C_5+\ldots+C_{n-1}=2^{n-1}\), if \(n\) is even
Options:
1. \(C_0+C_2+C_4+\ldots+C_n=2^{n-1}\), if \(n\) is even
2. \(C_1+C_3+C_5+\ldots+C_{n-1}=2^{n-1}\), if \(n\) is even
Solution:
2142 Upvotes
Verified Answer
The correct answer is:
Both 1 and 2 are true
Since,
\((1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots \ldots+C_n x^n\)
On putting \(x=1\) and \(x=-1\) respectively, we get
\(\begin{aligned}
2^n & =C_0+C_1+C_2+C_3+\ldots \ldots+C_n \\
0 & =C_0-C_1+C_2-C_3+\ldots \ldots+(-1)^n C_n
\end{aligned}\)
and \(0=C_0-C_1+C_2-C_3+\ldots \ldots+(-1)^n C_n\)
If \(n\) is even , then
\(C_0+C_1+C_2+C_3+\ldots \ldots+C_n=2^n\)...(i)
\(\text {and } C_0-C_1+C_2-C_3+\ldots \ldots+C_n=0\)...(ii)
By adding and subtracting Eqs. (i) and (ii), we get
\(C_0+C_2+C_4+\ldots \ldots+C_n=2^{n-1}\)
and \(\quad C_1+C_3+C_5+\ldots \ldots+C_{n-1}=2^{n-1}\)
\((1+x)^n=C_0+C_1 x+C_2 x^2+C_3 x^3+\ldots \ldots+C_n x^n\)
On putting \(x=1\) and \(x=-1\) respectively, we get
\(\begin{aligned}
2^n & =C_0+C_1+C_2+C_3+\ldots \ldots+C_n \\
0 & =C_0-C_1+C_2-C_3+\ldots \ldots+(-1)^n C_n
\end{aligned}\)
and \(0=C_0-C_1+C_2-C_3+\ldots \ldots+(-1)^n C_n\)
If \(n\) is even , then
\(C_0+C_1+C_2+C_3+\ldots \ldots+C_n=2^n\)...(i)
\(\text {and } C_0-C_1+C_2-C_3+\ldots \ldots+C_n=0\)...(ii)
By adding and subtracting Eqs. (i) and (ii), we get
\(C_0+C_2+C_4+\ldots \ldots+C_n=2^{n-1}\)
and \(\quad C_1+C_3+C_5+\ldots \ldots+C_{n-1}=2^{n-1}\)
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