Question

If $n$ is an odd integer and $a,b,c$ are distinct, find the number of distinct terms in the expansion of ${(a+b+c)}^n+{(a+b-c)}^n$

• A. $n+1$
• B. $\frac{(n+1)(n+3)}{4}$
• C. $(n+1)\left(n\right)$
• D. $\frac{(n+1)(n-1)}{4}$

Answer 1

Answer: (B)

$\frac{(n+1)(n+3)}{4}$

Let $n=3$
$(a+b-c{)}^3$
$=(a+b{)}^3-c^3-3c(a+b\left)\right[a+b-c]$
$=a^3+b^3-c^3+3a^{2}b+3a{b}^2+3a{c}^2+3b{c}^2-3c(a+b{)}^2$
$=a^3+b^3-c^3+3a^{2}b+3a{b}^2+3a{c}^2+3b{c}^2-3a^{2}c-3b{c}^2-6abc$
And
$(a+b+c{)}^3$
$=a^3+b^3+c^3+3a^{2}b+3a{b}^2+3a{c}^2+3b{c}^2+3a^{2}c+3b{c}^2+6abc$
Adding both, we get
$(a+b+c{)}^3+(a+b-c{)}^3=2[a^3+b^3+3a^{2}b+3a{b}^2+3a{c}^2+3b{c}^2]$
Hence 6 dissimilar terms
$=\frac{4.6}{4}$
$=\frac{(3+1)(3+3)}{4}$.
Here $n=3$
Hence the correct option is $\frac{(n+1)(n+3)}{4}$.