displaystyle ∑ r = 0n ((r2/r + 1))nCr is equal to

Question

displaystyle ∑ r = 0n ((r2/r + 1))nCr is equal to (A) (2n - 1 (n2 + n + 2) - 1/(n + 1)) (B) (2n - 1 (n2 - n - 2) + 1/(n + 1)) (C) (2n - 1 (n2 - n + 2) - 1/(n + 1)) (D) (2n - 1 (n2 + n - 2) + 1/(n + 1))

Tardigrade Admin · Accepted Answer

displaystyle ∑ r = 0n((r2/r + 1))nCr= displaystyle ∑ r = 0n((r2 - 1 + 1/r + 1))nCr = displaystyle ∑ r = 0n((r - 1) + (1/r + 1))nCr= displaystyle ∑ r = 0nr(⋅ )nCr-( displaystyle ∑ r = 0n)nCr+ displaystyle ∑ r = 0n((1/r + 1))nCr (1/r + 1)n Cr = (1/n + 1)n + 1 Cr + 1 and r ⋅ n Cr = n ⋅ n - 1 Cr - 1 = displaystyle ∑ r = 1nn⋅ n - 1Cr - 1-2n+ displaystyle ∑ r = 0n(1/n + 1)⋅ n + 1Cr + 1 =n n - 1 C0 +n - 1 C1 + . . . +n - 1 Cn - 1 -2n+(1/n + 1) n + 1 C1 +n + 1 C2 + . . . +n + 1 Cn + 1 =n⋅ 2n - 1-2n+(1/n + 1)(2n + 1 - 1) =(1/(n + 1)) 2n + 1 - 1 + 2n - 1 (n - 2) (n + 1) =(2n - 1 (n2 - n + 2) - 1/(n + 1))

Q. $r = 0 \sum n (\frac{r ^{2}}{r + 1})^{n} C_{r}$ is equal to

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

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

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

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

Q. r=0∑n​(r+1r2​)nCr​ is equal to

(n+1)2n−1(n2+n+2)−1​

(n+1)2n−1(n2−n−2)+1​

(n+1)2n−1(n2−n+2)−1​

(n+1)2n−1(n2+n−2)+1​

Q. $r = 0 \sum n (\frac{r ^{2}}{r + 1})^{n} C_{r}$ is equal to

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

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

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

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