- r-0 n-1 n C r n C r - n C r-1 equals

The correct option is D n(n 1)2(n+1) ∑_r=0^n 1^nC_r^nC_r+^nC_r 1 ⇒∑_r=0^n 1^nC_r^n+1C_r⇒n!(n r)!r!(n+1)!(n+1 r!)r! ⇒∑_r=0^n 11n+11(n+1 r) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Method of Difference

∑ r=0 n-1 n C...

Question

$\sum_{r = 0}^{n - 1} \frac{n_{C_{r}}}{n_{C_{r}} + n_{C r + 1}}$ equals

\frac{n}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n + 1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n (n + 1)}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{n (n - 1)}{2 (n + 1)}$
$n - 1 \sum r = 0 \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r - 1}}$
$\Rightarrow n - 1 \sum r = 0 \frac{^{n} C_{r}}{^{n + 1} C_{r}} \Rightarrow \frac{n! (n - r)!}{r!} \frac{(n + 1)! (n + 1 - r!)}{r!}$
$\Rightarrow n - 1 \sum r = 0 \frac{1}{n + 1} \frac{1}{(n + 1 - r)}$
$\Rightarrow \frac{1}{n + 1} n - 1 \sum r = 0 \frac{1}{(n + 1 - r)}$
$\Rightarrow \frac{1}{(n + 1)} \frac{(n - 1) n}{2} \Rightarrow \frac{n (n - 1)}{2 (n + 1)}$

Suggest Corrections

Similar questions

Let r and n be positive integers such that $1 \leq r \leq n .$ Then prove the following :

(i) $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
(ii) $n^{n - 1} C_{r - 1} = (n - r + 1)^{n} C_{r - 1}$
(iii) $\frac{^{n} C_{r}}{^{n - 1} C_{r - 1}} = \frac{n}{r}$
(iv) $^{n} C_{r} + 2^{n} C_{r - 1} +^{n} C_{r - 2} =^{n + 2} C_{r}$

Q. The value of

n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}}

Q. Sum of the series

n \sum r = 0 (- 1)^{r}^{n} C_{r} \cdot [i^{5 r} + i^{6 r} + i^{7 r} + i^{8 r}],

Q. The value of

n - 1 \sum r = 0 \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}}

is equals to ..............

Q. If

\frac{^{n} C_{r}}{^{n} C_{r + 1}} = \frac{1}{2}

and

\frac{^{n} C_{r + 1}}{^{n} C_{r + 2}} = \frac{2}{3}

; determine the values of

n

and

r

Join BYJU'S Learning Program

∑n−1r=0nCrnCr+nCr+1 equals

The correct option is D n(n−1)2(n+1)n−1∑r=0nCrnCr+nCr−1⇒n−1∑r=0nCrn+1Cr⇒n!(n−r)!r!(n+1)!(n+1−r!)r!⇒n−1∑r=01n+11(n+1−r)⇒1n+1n−1∑r=01(n+1−r)⇒1(n+1)(n−1)n2⇒n(n−1)2(n+1)

$\sum_{r = 0}^{n - 1} \frac{n_{C_{r}}}{n_{C_{r}} + n_{C r + 1}}$ equals