If C0,C1, C2, ....,Cn are binomial coefficients of order n, then the value of (C1/2)+(C3/4)+(C5/6)+...=

Question

If C0,C1, C2, ....,Cn are binomial coefficients of order n, then the value of (C1/2)+(C3/4)+(C5/6)+...= (A) (2n +1/n+1) (B) (2n -1/n+1) (C) (2n +1/n -1) (D) (2n /n+1)

Tardigrade Admin · Accepted Answer

(C1/2)+(C3/4)+(C5/6)+...=(n/1⋅ 2) +( n(n-1)(n-2)/1⋅ 2⋅ 3⋅ 4) (+n(n-1)(n-2)(n-3)(n-4)/1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6)+..... = (1/(n+1))[(n(n+1)/2!)+(n(n-1)(n-2)(n+1)/4!)+...] =(1/n+1)[n+1C2+ n+1C4 +......] =(1/n+1) [2n+1-1-1] =(2n -1/n+1)

Q. If $C_{0}, C_{1}, C_{2}, ...., C_{n}$ are binomial coefficients of order n, then the value of $\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + ... =$

$\frac{2 ^{n} + 1}{n + 1}$

$\frac{2 ^{n} - 1}{n + 1}$

$\frac{2 ^{n} + 1}{n - 1}$

$\frac{2 ^{n}}{n + 1}$

Q. If C0​,C1​,C2​,....,Cn​ are binomial coefficients of order n, then the value of 2C1​​+4C3​​+6C5​​+...=

n+12n+1​

n+12n−1​

n−12n+1​

n+12n​

2C1​​+4C3​​+6C5​​+...=1⋅2n​ +1⋅2⋅3⋅4n(n−1)(n−2)​1⋅2⋅3⋅4⋅5⋅6+n(n−1)(n−2)(n−3)(n−4)​+..... =(n+1)1​[2!n(n+1)​+4!n(n−1)(n−2)(n+1)​+...] =n+11​[n+1C2​+n+1C4​+......] =n+11​[2n+1−1−1]=n+12n−1​

Q. If $C_{0}, C_{1}, C_{2}, ...., C_{n}$ are binomial coefficients of order n, then the value of $\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + ... =$

$\frac{2 ^{n} + 1}{n + 1}$

$\frac{2 ^{n} - 1}{n + 1}$

$\frac{2 ^{n} + 1}{n - 1}$

$\frac{2 ^{n}}{n + 1}$