Question

Number of integral terms in the expansion of $(\sqrt{6}+\sqrt{7}{)}^{32}$ is equal to:

• A. $16$
• B. $17$
• C. $18$
• D. $20$

Answer 1

The correct option is B $17$

General term in the expansion ${(\sqrt{6}+\sqrt{7})}^{32}$ is

$T_{r+1}{=}^{32}{C}_r.6^{\frac{32-r}{2}}.7^{\frac{r}{2}}$

$r$ can be 0 to 32

Number of integral terms $=\frac{32}{2}+1=16+1=17$ (for r=0 and all even r)