Let the sum of the coefficients of the first three terms in the expansion of ( x - 3 / x^2 )^n,x ne 0. n in mathbbN, be 376. Then the coefficient of x^4 is __________.

S=1-3 n+(9 n(n-1) / 2)=376 3 n^2-5 n-250=0 \ n=10, (-25 / 3) (Rejected) \ T_r+1= ^n C_r * x^n-r(frac-3x^2)^r \ = ^n C_r x x^n-3 r(-3)^r \ = ^10 C_r x^10-3 r(-3)^r Here r=2 Required coefficient = ^10 C_2(-3)^2 \ =45 x 9 \ =405

Let the sum of the coefficients of the first three term... | null (Mathematics)

S=1-3 n+\frac{9 n(n-1)}{2}=376

\begin{aligned} & 3 n^{2}-5 n-250=0 \\\\ & n=10, \frac{-25}{3} \text { (Rejected) } \\\\ & T_{r+1}={ }^{n} C_{r} \cdot x^{n-r}\left(\frac{-3}{x^{2}}\right)^{r} \\\\ & ={ }^{n} C_{r} x x^{n-3 r}(-3)^{r} \\\\ & ={ }^{10} C_{r} x^{10-3 r}(-3)^{r} \end{aligned}

Here

r=2

\begin{aligned} \text {Required coefficient } & ={ }^{10} \mathrm{C}_{2}(-3)^{2} \\\\ & =45 \times 9 \\\\ & =405 \end{aligned}

Explanation

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