Question

If the coefficients of $(2r+1)$th term and $(r+1)$th in the expansion of ${(1+x)}^{42}$ are equal then $r$ can be

• A. $12$
• B. $14$
• C. $16$
• D. $20$

Answer 1

The correct option is B $14$

In the expansion of $(1+x{)}^{42}$, $(r+1{)}^{th}$ term is given by $T_{r+1}=(\frac{42}{r})x^r$

=> coefficient of this term is $(\frac{42}{r})$

SImilarly, coefficient of $(2r+1{)}^{th}$ term is $(\frac{42}{2r})$

We need to solve for

$(\frac{42}{r})=(\frac{42}{2r})$

We know the property, that $(\frac{n}{r})=(\frac{n}{n-r})$

=> $(\frac{42}{r})=(\frac{42}{42-2r})$

=> $r=42-2r$

=>$r=14$