If $C_{r} =^{n} C_{r}$ and $(C_{0} + C_{1}) (C_{1} + C_{2}) . . . (C_{n - 1} + C_{n}) = k$ $(C_{0} C_{1} C_{2} . . . C_{n})$ then the value of $k$ equals

\frac{(n + 1)^{n + 1}}{n!}

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

\frac{(n + 1)^{n}}{n . n!}

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

\frac{(n)^{n}}{n!}

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

\frac{(n + 1)^{n}}{n!}

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

Open in App

Solution

The correct option is B $\frac{(n + 1)^{n}}{n!}$
$^{n} C_{r + 1} +^{n} C_{r}$
$=^{n + 1} C_{r + 1}$
Hence simplifying the terms, we get
$^{n + 1} C_{1} .^{n + 1} C_{2} . . . .^{n + 1} C_{n}$
Now
$^{n + 1} C_{1}$
$= \frac{(n + 1)!}{n! .1!}$
$= \frac{(n + 1)}{n}^{n} C_{1}$
Similarly
$^{n + 1} C_{2}$
$= \frac{(n + 1)!}{(n - 1)! .2!}$
$= \frac{(n + 1)}{(n - 1)}^{n} C_{2}$
Hence substituting, in the above expression, we get
$\frac{(n + 1)^{n}}{n!} (^{n} C_{0} .^{n} C_{1} . . .^{n} C_{n})$
Comparing coefficients, we get
$K = \frac{(n + 1)^{n}}{n!}$

Suggest Corrections

Similar questions

If $C_{r} =^{n} C_{r} a n d (C_{0} + C_{1}) (C_{1} + C_{2}) . . . . (C_{n - 1} + C_{n}) = k \frac{(n + 1)^{n}}{n!}$ , then the value of k is :

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}

(C_{0} + C_{1}) (C_{1} + C_{2}) \dots (C_{n - 1} + C_{n}) = k \cdot C_{1} C_{2} C_{3} \dots C_{n}

k

(C_{0} + C_{1}) (C_{1} + C_{2}) (C_{2} + C_{3}) (C_{3} + C_{4}) . . . . . (C_{n - 1} + C_{n}) = \frac{C_{0} C_{1} C_{2} . . . . C_{n - 1} {(n + 1)}^{n}}{n!}

(C_{0} + C_{1}) (C_{1} + C_{2}) . . . . . (C_{n - 1} + C_{n})

= \frac{(n + 1)^{n}}{n!} C_{1} C_{2} C_{3} . . . . . C_{n}

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

{(1 + x)}^{n}

(c_{0} + c_{1}) (c_{1} + c_{2}) . . . . . . . (c_{n - 1} + c_{n}) = \frac{c_{1} c_{2} . . . c_{n} {(n + 1)}^{n}}{| n - -}

Join BYJU'S Learning Program