C0-32-C1- C2-274-C3-3nn-1-Cn-4n-1-13n-1

The correct option is B Falseclassfollow (1+x)^n=^n C_0+^n C_1 x+^n C_2 x^2+…+^n C_n x^n Integrate on both sides from 0 to? 4^n+1/n+1 1=^n C_0(3)+^n C_ ...

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

Byju's Answer

Standard XII

Mathematics

Integration to Solve Modified Sum of Binomial Coefficients

C0+32.C1+ C2+...

Question

$C_{0} + \frac{3}{2} . C_{1} + C_{2} + \frac{27}{4} . C_{3} + + \frac{3^{n}}{n + 1} . C_{n} = \frac{4^{n + 1} + 1}{3 (n + 1)}$

True

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

False

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

Open in App

Solution

The correct option is B False
classfollow
$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}$
Integrate on both sides from 0 to?
$\frac{4^{n + 1}}{n + 1} - 1 =^{n} C_{0} (3) +^{n} C_{1} (\frac{3)}{2}^{2} +^{n} C_{2} \frac{3^{3}}{3} + \dots$
$t - \frac{n_{n} 3^{n + 1}}{n + 1}$
$\frac{1}{3} [\frac{4^{n + 1} - 1}{n + 1}] = C_{0} + \frac{3}{2} C_{1} + 3 C_{2} + \dots + \frac{3^{n}}{n + 1} C_{n}$
So, option is false

Suggest Corrections

Similar questions

C_{0} + \frac{3}{2} . C_{1} + \frac{9}{3} . C_{2} + \frac{27}{4} . C_{3} + . . . . . . . . . . . + \frac{3^{n}}{n + 1} . C_{n} = \frac{4^{n + 1} - 1}{3 (n + 1)}

.Is it true ?If true enter 1 else 0.

Q. Prove that

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

Q. The value of

2 (^{n} C_{0}) + \frac{3}{2} (^{n} C_{1}) + \frac{4}{3} (^{n} C_{2}) + \frac{5}{4} (^{n} C_{3}) . . .

is:

\frac{C_{1}}{C_{0}} + 2. \frac{C_{2}}{C_{1}} + 3. \frac{C_{3}}{C_{2}} + . . . . . + n . \frac{C_{n}}{C_{n - 1}}

c_{0}, c_{1}, c_{2}

denotes coefficents expansion of

(1 + x)^{n}

, then

c_{1} + c_{1} c_{2} + c_{2} c_{3} + . . . . . . . c_{n - 1} c_{n} = \frac{(2 n)!}{(n + 1)! (n - 1)!}

Join BYJU'S Learning Program

C0+32.C1+C2+274.C3++3nn+1.Cn=4n+1+13(n+1)

The correct option is B Falseclassfollow(1+x)n=nC0+nC1x+nC2x2+…+nCnxnIntegrate on both sides from 0 to?4n+1n+1−1=nC0(3)+nC1(3)22+nC2333+⋯t−nn3n+1n+113[4n+1−1n+1]=C0+32C1+3C2+…+3nn+1CnSo, option is false

$C_{0} + \frac{3}{2} . C_{1} + C_{2} + \frac{27}{4} . C_{3} + + \frac{3^{n}}{n + 1} . C_{n} = \frac{4^{n + 1} + 1}{3 (n + 1)}$