2- C 0 - 2 2 - C 1 - 2 - 2 3 - C 2 - 3 - 2 n-1 - C n - n-1 -

The correct option is C 3^ n +1/ n+1 We know that ( 1+x ) #160; ^ n =^ n C _ 0 +^ n C _ 1 x+^ n C _ 2 x ^ 2 +....+....^ n C _ n x ^ n Integra ...

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

Combination

2. C 0 + 2 2 ...

Question

$2. C_{0} + 2^{2} . \frac{C_{1}}{2} + 2^{3} . \frac{C_{2}}{3} + . . . + 2^{n + 1} . \frac{C_{n}}{n + 1} =$

\frac{3^{n + 1} - 1}{2 (n + 1)}

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

\frac{3^{n + 1} - 1}{n + 1}

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

\frac{3^{n} - 1}{n + 1}

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

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

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

Open in App

Solution

Suggest Corrections

Similar questions

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)}

S = \frac{2}{1} {^{n} C}_{0} + \frac{2^{2}}{2} {^{n} C}_{1} + \frac{2^{3}}{3} {^{n} C}_{2} + . . . . . + \frac{2^{n + 1}}{n + 1} {^{n} C}_{n}

S

S = \frac{2}{1}^{n} C_{0} + \frac{2^{2}}{2}^{n} C_{1} + \frac{2^{3}}{3}^{n} C_{2} + \dots + \frac{2^{n + 1}}{n + 1}^{n} C_{n}

S

2 C_{0} + \frac{C_{1}}{2} \cdot 2^{2} + \frac{C_{2}}{3} \cdot 2^{3} + \frac{C_{3}}{4} \cdot 2^{4} + \dots + \frac{C_{n}}{n + 1} \cdot 2^{n + 1}

(C_{r} =^{n} C_{r})

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

Join BYJU'S Learning Program