^{15}C_3 + ^{15}C_5 + ldots + ^{15}C_{15} = ? | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that the sum of binomial coefficients with odd indices is given by: $^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \ldots = 2^{n-1}$. For $n = 15$, we hav

Vedclass · Accepted Answer

$2^{14} - 15$

Vedclass · Answer

(B) We know that the sum of binomial coefficients with odd indices is given by: $^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \ldots = 2^{n-1}$. For $n = 15$, we have: $^{15}C_1 + ^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = 2^{15-1} = 2^{14}$. We need to find the sum $^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15}$. This is equal to $2^{14} - ^{15}C_1$. Since $^{15}C_1 = 15$, the sum is $2^{14} - 15$.

$^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = ?$

Similar Questions

The value of ${ }^{10} C_{1}+{ }^{10} C_{2}+{ }^{10} C_{3}+\ldots+{ }^{10} C_{9}$ is

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$,then $C_0 + 2 C_1 + 3 C_2 + \ldots + (n+1) C_n$ is equal to

If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$, then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + \dots + a_{2n}$ is equal to

The sum of the last eight coefficients in the expansion of $(1 + x)^{15}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module