The value of C0-1-3-C1-2-3-C2-3-3-C3-4-3-1nCn-n-1 -3 isA- 3-n-1B- n -1-3C- 1-3n-1D- None of these

The correct option is C 1/3(n+1) (1 + x)^n = C_0 + C_1x +C_2x^2 +⋯ + C_nx^n On integrating from 1 to 0 C_0 –C_1/2+C_2/3⋯ + (–1)^n C_n/n+1=1/n+1 Hence require ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

General Term of Binomial Expansion

The value of ...

Question

The value of $\frac{C_{0}}{1.3} - \frac{C_{1}}{2.3} + \frac{C_{2}}{3.3} - \frac{C_{3}}{4.3} + \dots + (- 1)^{n} \frac{C_{n}}{(n + 1) .3}$ is

$\frac{3}{n + 1}$

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

$\frac{n + 1}{3}$

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

$\frac{1}{3 (n + 1)}$

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

None of these

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

Open in App

Solution

The correct option is C
$\frac{1}{3 (n + 1)}$

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$
On integrating from - 1 to 0
$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} \dots + (- 1)^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}$
Hence required = $\frac{1}{3 (n + 1)}$

Suggest Corrections

The value of C01.3−C12.3+C23.3−C34.3+⋯+(−1)nCn(n+1).3 is

The correct option is C 13(n+1) (1+x)n=C0+C1x+C2x2+⋯+Cnxn On integrating from - 1 to 0 C0–C12+C23⋯+(–1)nCnn+1=1n+1 Hence required = 13(n+1)

The value of $\frac{C_{0}}{1.3} - \frac{C_{1}}{2.3} + \frac{C_{2}}{3.3} - \frac{C_{3}}{4.3} + \dots + (- 1)^{n} \frac{C_{n}}{(n + 1) .3}$ is

The correct option is C
$\frac{1}{3 (n + 1)}$

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$
On integrating from - 1 to 0
$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} \dots + (- 1)^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}$
Hence required = $\frac{1}{3 (n + 1)}$