Question

The coefficient of $x^5$ in the expansion of $(1+x{)}^{21}+(1+x{)}^{22}+\dots \dots ..+(1+x{)}^{30}$ is

• A. ${}^{51}{C}_5$
• B. ${}^9{C}_5$
• C. ${}^{31}{C_6-}^{21}{C}_6$
• D. ${}^{30}{C_5+}^{20}{C}_5$

Answer 1

Answer: (C)

${}^{31}{C_6-}^{21}{C}_6$

Let
$S=(1+x{)}^{21}+(1+x{)}^{22}+(1+x{)}^{23}+...(1+x{)}^{30}$ ...(i)

$(1+x)S=(1+x{)}^{22}+(1+x{)}^{23}...(1+x{)}^{30}+(1+x{)}^{31}$ ...(ii)

Subtracting $\left(i\right)$ from $\left(ii\right)$, we get

$xS=(1+x{)}^{31}-(1+x{)}^{21}$

$S=\frac{(1+x{)}^{31}}{x}-\frac{(1+x{)}^{21}}{x}$

Coefficient of $x^r$

$={}^{31}{C}_r{x}^{r-1}-{}^{21}{C}_r{x}^{r-1}$

$=({}^{31}{C}_r-{}^{21}{C}_r)x^{r-1}$ ... $\left(a\right)$

Therefore the coefficient of $x^5$ implies

$r-1=5$

$r=6$

Substituting in $\left(a\right)$, we get coefficient of $x^5$

$=({}^{31}{C}_6-{}^{21}{C}_6)$