Question

Coefficient of the middle term in the expansion of the binomial $(1+x{)}^{40}$ is equal to?

• A. $\frac{2^{20}\cdot (1\cdot 3\cdot \mathrm{5......39})}{\left(20\right)!}$
• B. The sum of the coefficients of the two middle terms in $(1+x{)}^{39}$
• C. $\frac{20\cdot 21\cdot \mathrm{22....40}}{\left(20\right)!}$
• D. ${(}^{20}{C}_0{)}^2+{(}^{20}{C}_1{)}^2+{(}^{20}{C}_2{)}^2+....+{(}^{20}{C}_{20}{)}^2$

Answer 1

Answer: (B), (D)

The correct options are
B The sum of the coefficients of the two middle terms in $(1+x{)}^{39}$
D ${(}^{20}{C}_0{)}^2+{(}^{20}{C}_1{)}^2+{(}^{20}{C}_2{)}^2+....+{(}^{20}{C}_{20}{)}^2$

In ${(1+x)}^{40}$ middle term is $21st$ i.e. ${\left(T_{20+1}\right)}^{th}$

$\left(T_{20+1}\right){=}^{40}{C}_{20}{x}^{20}$

Co-efficient is ${}^{40}{C}_{20}=\frac{40!\times 20!}{20!}=40!$

and middle term of $(1+x{)}^{39}$ are ${}^{39}{C}_{20}$ and ${}^{39}{C}_{19}$

=${}^{39}{C}_{20}$ $+$${}^{39}{C}_{19}{=}^{40}{C}_{20}=40!$