If Cr-n Cr and C0-C1C1-C2 -Cn-1-Cn-k n-1n-n - then the value of k is -A- C 0 C 1 C 2- C nB- C12 C22- Cn2C- C 1- C 2- C nD- None of these-

The correct option is A ( C_0 + C_1 ) ( C_1 + C_2 ) ( C_2 + C_3) .... ( C_n 1 + C_n ) nbsp; = C_1C_2 ...C_n ( 1 + C_0/C_1) ( 1 + C_1/C_2) .... (1 + C_n 1/C_ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Continuity of a Function

If Cr=n Cr an...

Question

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 :

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

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

Open in App

Solution

The correct option is A

$(C_{0} + C_{1}) (C_{1} + C_{2}) (C_{2} + C_{3}) . . . . (C_{n - 1} + C_{n})$

= $C_{1} C_{2} . . . C_{n} (1 + \frac{C_{0}}{C_{1}}) (1 + \frac{C_{1}}{C_{2}}) . . . . (1 + \frac{C_{n - 1}}{C_{n}})$

= $C_{1} C_{2} . . . . C_{n} (\frac{n + 1}{n}) (\frac{n + 1}{n}) . . . . . (\frac{1 + n}{1})$

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

= $C_{0} C_{1} C_{2} . . . . C_{n} . \frac{(n + 1)^{n}}{n!}$ [ $C_{0} = 1]$

$∴$ k = $C_{0} C_{1} C_{2} . . . . . . C_{n} .$

Suggest Corrections

Similar questions

Q. 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

Q. If

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

denote the coefficients in the expansion of

{(1 + x)}^{n}

, prove that

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

Q. Prove that

(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!}

Q. If

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

, and

(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}

then

k

is equal to

If $^{n} C_{r} = C_{r}$ then $C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . + (C_{0} + C_{1} + . . . . . + C_{n})$ =

Join BYJU'S Learning Program

If Cr=nCr and (C0+C1)(C1+C2)....(Cn−1+Cn)=k(n+1)nn! , then the value of k is :

The correct option is A (C0+C1)(C1+C2)(C2+C3)....(Cn−1+Cn) = C1C2...Cn(1+C0C1)(1+C1C2)....(1+Cn−1Cn) = C1C2....Cn(n+1n)(n+1n).....(1+n1) = C1C2.....Cn.(n+1)nn! = C0C1C2....Cn.(n+1)nn! [ C0=1] ∴ k = C0C1C2......Cn.

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 :