displaystyle∑k=0n (nCk/k+1) =

Question

displaystyle∑k=0n (nCk/k+1) =  (A)  (2n -1/n+1) (B)  (2n-1 -1/n - 1) (C)  (2n + 1 -1/n + 1) (D)  (2n -1/n)

Tardigrade Admin · Accepted Answer

displaystyle∑k=0n (nCk/k+1) = displaystyle∑nk=0 (n!/(k+1)k! (n -k)!) = displaystyle∑k=0n (n!(n+1)/(k+1)!(n-k)!(n+1)) = (1/n+1) displaystyle∑ k=0n ((n+1)!/(k+1)!(n+1-k-1)!) = (1/n+1) displaystyle∑k = 0n n+1Ck +1 = (1/n+1) [ 2n+1 - 1]

Q. $k = 0 \sum n \frac{^{n} C _{k}}{k + 1} =$

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

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

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

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

Q. k=0∑n​k+1nCk​​=

n+12n−1​

n−12n−1−1​

n+12n+1−1​

n2n−1​

k=0∑n​k+1nCk​​=k=0∑n​(k+1)k!(n−k)!n!​ =k=0∑n​(k+1)!(n−k)!(n+1)n!(n+1)​ =n+11​k=0∑n​(k+1)!(n+1−k−1)!(n+1)!​ =n+11​k=0∑n​n+1Ck+1​=n+11​[2n+1−1]

Q. $k = 0 \sum n \frac{^{n} C _{k}}{k + 1} =$

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

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

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

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