Q119 / 180 · COMEDK 2021 · Mathematics
The coefficient of $x^{10}$ in the expansion of $1+(1+x)+\ldots+(1+x)^{20}$ is
- A${ }^{19} C_{9}$
- B${ }^{20} C_{10}$
- Correct answerC${ }^{21} C_{11}$
- D${ }^{22} C_{12}$
Solution
The given expression is a geometric series with first term $a = 1$, common ratio $r = (1+x)$, and number of terms $n = 21$.
The sum of the series is $S = a \dfrac{r^n - 1}{r - 1} = 1 \times \dfrac{(1+x)^{21} - 1}{(1+x) - 1} = \dfrac{(1+x)^{21} - 1}{x}$.
We need the coefficient of $x^{10}$ in the expansion of $S = \dfrac{(1+x)^{21} - 1}{x}$.
This is equivalent to finding the coefficient of $x^{11}$ in the expansion of $(1+x)^{21} - 1$.
The general term in the expansion of $(1+x)^{21}$ is given by $^{21}C_{r} x^{r}$.
For $r = 11$, the term is $^{21}C_{11} x^{11}$.
Since the constant term $-1$ does not affect the coefficient of $x^{11}$, the coefficient of $x^{11}$ in $(1+x)^{21} - 1$ is $^{21}C_{11}$.
Therefore, the coefficient of $x^{10}$ in the original expression is $^{21}C_{11}$.
Answer: ${ }^{21} C_{11}$
The sum of the series is $S = a \dfrac{r^n - 1}{r - 1} = 1 \times \dfrac{(1+x)^{21} - 1}{(1+x) - 1} = \dfrac{(1+x)^{21} - 1}{x}$.
We need the coefficient of $x^{10}$ in the expansion of $S = \dfrac{(1+x)^{21} - 1}{x}$.
This is equivalent to finding the coefficient of $x^{11}$ in the expansion of $(1+x)^{21} - 1$.
The general term in the expansion of $(1+x)^{21}$ is given by $^{21}C_{r} x^{r}$.
For $r = 11$, the term is $^{21}C_{11} x^{11}$.
Since the constant term $-1$ does not affect the coefficient of $x^{11}$, the coefficient of $x^{11}$ in $(1+x)^{21} - 1$ is $^{21}C_{11}$.
Therefore, the coefficient of $x^{10}$ in the original expression is $^{21}C_{11}$.
Answer: ${ }^{21} C_{11}$
Hence, the correct answer is option: 'C'.