The number of terms in the expansion of a-b-cn are-A- n-1n-2-4B- n-1n-1-2C- n-1n-1-4D- n-1n-2-2

The correct option is D nbsp; nbsp; nbsp; nbsp; ...

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

General Term of Binomial Expansion

The number of...

Question

The number of terms in the expansion of (a + b + c)ⁿ are:

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

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

Suggest Corrections

Similar questions

The number of terms in the expansion of $(a + b + c)^{n}$ will be

(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}

n

(1 \times 2 \times 3) + (2 \times 3 \times 4) + (3 \times 4 \times 5) + \dots

n

n^{n + 1} - n {(n - 1)}^{n + 1} + \frac{n (n - 1)}{2!} {(n - 2)}^{n + 1} - \dots = \frac{1}{2} n (n + 1)!

n^{n} - (n + 1) {(n - 1)}^{n} + \frac{(n + 1) n}{2!} {(n - 2)}^{n} - \dots = 1

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

{(n + p)}^{n} - n {(n + p - 1)}^{n} + \frac{n (n - 1)}{2!} {(n + p - 2)}^{n} - \dots = n!

n + 1

n^{th}

{log}_{e} (\frac{4}{3})

Join BYJU'S Learning Program