The number of distinct terms in the expansion of (x3+1+(1/x3))n ; x ∈ R+ and n ∈ N is

Question

The number of distinct terms in the expansion of (x3+1+(1/x3))n ; x ∈ R+ and n ∈ N is (A) 2 n (B) 3 n (C) 2 n+1 (D) 3 n+1

Tardigrade Admin · Accepted Answer

(x3+1+(1/x3))n=[1+(x3+(1/x3))]n = n C0+ n C1(x3+(1/x3))+ ldots+ n Cn(x3+(1/x3))n All the terms are distinct with powers (x3)0,(x3),(x3)2, ldots(x3) n,(x3)-n, ldots(x3)-1. Hence, (2 n+1) terms.

Q. The number of distinct terms in the expansion of $(x^{3} + 1 + \frac{1}{x ^{3}})^{n}; x \in R^{+}$ and $n \in N$ is

$2 n$

$3 n$

$2 n + 1$

$3 n + 1$

Q. The number of distinct terms in the expansion of (x3+1+x31​)n;x∈R+ and n∈N is

2n

3n

2n+1

3n+1

(x3+1+x31​)n=[1+(x3+x31​)]n =nC0​+nC1​(x3+x31​)+…+nCn​(x3+x31​)n All the terms are distinct with powers (x3)0,(x3),(x3)2,…(x3) n,(x3)−n,…(x3)−1. Hence, (2n+1) terms.

Q. The number of distinct terms in the expansion of $(x^{3} + 1 + \frac{1}{x ^{3}})^{n}; x \in R^{+}$ and $n \in N$ is

$2 n$

$3 n$

$2 n + 1$

$3 n + 1$