Question

$^{15}C_{3} +\,^{15}C_{5} + .. ...+\,^{15}C_{15} $ will be equal to

• A. $2^{14}$
• B. $2^{14} - 15$
• C. $2^{14} + 15$
• D. $2^{14} - 1$

Answer 1

The Correct Option is B

We know,
${ }^{15} C_{1}+{ }^{15} C_{3}+{ }^{15} C_{5}+\ldots+{ }^{15} C_{15}=2^{15-1}$
$\therefore{ }^{15} C_{3}+{ }^{15} C_{5}+\ldots+{ }^{15} C_{15}=2^{14}-15$

Answer 2

To find the value of \( ^{15}C_{3} + ^{15}C_{5} + \ldots + ^{15}C_{15} \), we can use the property of binomial coefficients that \( ^nC_r = ^nC_{n-r} \). Therefore, the given expression is equal to:
\[^{15}C_{3} + ^{15}C_{5} + \ldots + ^{15}C_{15} = \left( ^{15}C_{0} + ^{15}C_{1} + \ldots + ^{15}C_{15} \right) - \left( ^{15}C_{0} + ^{15}C_{1} + \ldots + ^{15}C_{14} \right)\]
Now, we know that the sum of all the binomial coefficients of a given degree is equal to \( 2^n \), where \( n \) is the degree. Therefore,
\[^{15}C_{0} + ^{15}C_{1} + \ldots + ^{15}C_{15} = 2^{15}\]
Similarly,
\[^{15}C_{0} + ^{15}C_{1} + \ldots + ^{15}C_{14} = 2^{15} - ^{15}C_{15}\]
Substituting these values back into the expression, we get:
\[^{15}C_{3} + ^{15}C_{5} + \ldots + ^{15}C_{15} = 2^{15} - (2^{15} - ^{15}C_{15}) = 2^{15} - 1 = 2^{14} - 15\]
Therefore, \( ^{15}C_{3} + ^{15}C_{5} + \ldots + ^{15}C_{15} = 2^{14} - 15 \).
So the correct Answer is Option (B)