Question

Statement I ${ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots \ldots=2^{n-1}$
Statement II ${ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots \ldots=2^{n-1}$

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (A)

In the binomial expression, we have
$(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots+{ }^n C_n b^n .....$(i)
The coefficients ${ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots .,{ }^n C_n$ are known as binomial or combinatorial coefficients.
Putting $a=b=1$ in (i), we get
${ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_n=2^n$
Thus, the sum of all binomial coefficients is equal to $2^n$.
Again , putting $a=1$ and $b=-1$ in Eq. (i), we get
${ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots={ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots$
Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to $\frac{2^n}{2}=2^{n-1}$.
$\Rightarrow{ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots={ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots=2^{n-1}$