Let the sum of the coefficients of the first three terms in the expansion of (x-3x^2)^n, x…

Question

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\dfrac{3}{x^2}ight)^n$, $x
e0,\ n\in N$, be 376. Then the coefficient of $x^4$ is\_\_\_

Answer: 405

Accepted Answer

Sum of coefficients of first three terms: ${}^nC_0-3\cdot{}^nC_1+9\cdot{}^nC_2=376$

$$1-3n+\frac{9n(n-1)}{2}=376\quad\Rightarrow\quad 3n^2-5n-250=0\quad\Rightarrow\quad(n-10)(3n+25)=0\quad\Rightarrow\quad n=10$$

General term: $T_{r+1}=\binom{10}{r}x^{10-r}(-3)^rx^{-2r}=\binom{10}{r}(-3)^rx^{10-3r}$

For $x^4$: $10-3r=4\Rightarrow r=2$

$$\binom{10}{2}(-3)^2=45\times9=405$$

Solutions · 1