Let the sum of the coefficients of the first three terms in the expansion of ( x - 3 \over x^2 )^n,x e 0.~n \in \mathbbN , be 376. Then the coefficient of x^4 is \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2023 · 24 Jan · Shift 2 · Q40 with Solution

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Correct answer: 405

For $\left(x-\frac{3}{x^2}\right)^n,$ the general term is

=\binom{n}{r}(-3)^r x^{n-r-2r} =\binom{n}{r}(-3)^r x^{n-3r}.$$ 2. **Find the first three terms and their coefficients** - First term $(r=0)$: $$T_1=x^n,$$ coefficient $=1$. - Second term $(r=1)$: $$T_2=\binom{n}{1}(-3)x^{n-3}=-3n\,x^{n-3},$$ coefficient $=-3n$. - Third term $(r=2)$: $$T_3=\binom{n}{2}9x^{n-6},$$ coefficient $=9\binom{n}{2}$. So the sum of coefficients of the first three terms is $$1-3n+9\binom{n}{2}=376.$$ 3. **Solve for $n$** Using $$\binom{n}{2}=\frac{n(n-1)}{2},$$ we get $$1-3n+9\cdot \frac{n(n-1)}{2}=376.$$ Multiply by $2$: $$2-6n+9n(n-1)=752.$$ $$2-6n+9n^2-9n=752.$$ $$9n^2-15n-750=0.$$ Divide by $3$: $$3n^2-5n-250=0.$$ Now solve: $$3n^2-5n-250=0.$$ Discriminant: $$D=(-5)^2-4(3)(-250)=25+3000=3025=55^2.$$ Thus, $$n=\frac{5\pm 55}{6}.$$ Since $n\in\mathbb N$, $$n=\frac{60}{6}=10.$$ 4. **Find the coefficient of $x^4$** The power of $x$ in the general term is $$n-3r=10-3r.$$ For the term containing $x^4$, $$10-3r=4 \implies 3r=6 \implies r=2.$$ So the coefficient of $x^4$ is the coefficient of the $(r=2)$ term: $$\binom{10}{2}(-3)^2=45\cdot 9=405.$$ 5. **Final answer** The coefficient of $x^4$ is $$\boxed{405}.$$

View written solutionFree

Correct answer: 405

For $\left(x-\frac{3}{x^2}\right)^n,$ the general term is

=\binom{n}{r}(-3)^r x^{n-r-2r} =\binom{n}{r}(-3)^r x^{n-3r}.$$ 2. **Find the first three terms and their coefficients** - First term $(r=0)$: $$T_1=x^n,$$ coefficient $=1$. - Second term $(r=1)$: $$T_2=\binom{n}{1}(-3)x^{n-3}=-3n\,x^{n-3},$$ coefficient $=-3n$. - Third term $(r=2)$: $$T_3=\binom{n}{2}9x^{n-6},$$ coefficient $=9\binom{n}{2}$. So the sum of coefficients of the first three terms is $$1-3n+9\binom{n}{2}=376.$$ 3. **Solve for $n$** Using $$\binom{n}{2}=\frac{n(n-1)}{2},$$ we get $$1-3n+9\cdot \frac{n(n-1)}{2}=376.$$ Multiply by $2$: $$2-6n+9n(n-1)=752.$$ $$2-6n+9n^2-9n=752.$$ $$9n^2-15n-750=0.$$ Divide by $3$: $$3n^2-5n-250=0.$$ Now solve: $$3n^2-5n-250=0.$$ Discriminant: $$D=(-5)^2-4(3)(-250)=25+3000=3025=55^2.$$ Thus, $$n=\frac{5\pm 55}{6}.$$ Since $n\in\mathbb N$, $$n=\frac{60}{6}=10.$$ 4. **Find the coefficient of $x^4$** The power of $x$ in the general term is $$n-3r=10-3r.$$ For the term containing $x^4$, $$10-3r=4 \implies 3r=6 \implies r=2.$$ So the coefficient of $x^4$ is the coefficient of the $(r=2)$ term: $$\binom{10}{2}(-3)^2=45\cdot 9=405.$$ 5. **Final answer** The coefficient of $x^4$ is $$\boxed{405}.$$