Q76 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
The coefficient of $x^{29}$ in the expansion of $\left(1-3 x+3 x^2-x^3\right)^{15}$ is
- Correct answerA$-{ }^{45} C_{16}$
- B${ }^{45} C_{28}$
- C${ }^{45} C_{30}$
- D${ }^{45} C_{29}$
Solution
The given expression is $\left(1 - 3x + 3x^2 - x^3\right)^{15}$.
Observe that $1 - 3x + 3x^2 - x^3 = (1 - x)^3$.
Substituting this into the expression, we get $((1 - x)^3)^{15} = (1 - x)^{45}$.
The general term in the binomial expansion of $(1 - x)^{n}$ is given by $^{n}C_{r} (1)^{n-r} (-x)^{r} = ^{n}C_{r} (-1)^{r} x^{r}$.
For the expansion of $(1 - x)^{45}$, the term containing $x^{29}$ corresponds to $r = 29$.
The coefficient is $^{45}C_{29} (-1)^{29} = -^{45}C_{29}$.
Since $^{45}C_{29} = ^{45}C_{45-29} = ^{45}C_{16}$, the coefficient is $-^{45}C_{16}$.
Answer: $-{ }^{45} C_{16}$
Observe that $1 - 3x + 3x^2 - x^3 = (1 - x)^3$.
Substituting this into the expression, we get $((1 - x)^3)^{15} = (1 - x)^{45}$.
The general term in the binomial expansion of $(1 - x)^{n}$ is given by $^{n}C_{r} (1)^{n-r} (-x)^{r} = ^{n}C_{r} (-1)^{r} x^{r}$.
For the expansion of $(1 - x)^{45}$, the term containing $x^{29}$ corresponds to $r = 29$.
The coefficient is $^{45}C_{29} (-1)^{29} = -^{45}C_{29}$.
Since $^{45}C_{29} = ^{45}C_{45-29} = ^{45}C_{16}$, the coefficient is $-^{45}C_{16}$.
Answer: $-{ }^{45} C_{16}$
Hence, the correct answer is option: 'A'.