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JEE Main 202324 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 24 January · Evening Shift · Mathematics · Binomial Theorem

Let he sum of the coefficient of first three terms in the expansion...

Let he sum of the coefficient of first three terms in the expansion of (x−3x2)n;x=0,n∈N{\left(x - \dfrac{3}{{x}^{2}}\right)}^{n}; x = 0, n \in N be 376376. Then, the coefficient of x4{x}^{4} is equal to:

Correct answer

0

Step-by-step solution

We know that general term in the expansion of (x−3x2){\left(x - \dfrac{3}{{x}^{2}}\right)}^{n}isgivenby,</p><p>is given by,</p><p>{T}_{r + 1} = Cr n {x}^{n - r}{\left(\dfrac{- 3}{{x}^{2}}\right)}^{r}</p><p></p><p>= {\left(- 1\right)}^{r} \times Cr n {3}^{r}{x}^{n - r - 2r}</p><p></p><p>{T}_{r + 1} = {\left(- 1\right)}^{r} \times Cr n {3}^{r}{x}^{n - 3r} . . . . . . . . . . . . . . . \left(1\right)</p><p>So,</p><p>So,{T}_{1} = {T}_{0 + 1} = C0 n {3}^{0}{x}^{n} = {x}^{n}</p><p></p><p>{T}_{2} = \left(- 1\right) \times C1 n {3}^{1}{x}^{n - 3}</p><p></p><p>{T}_{3} = C2 n {3}^{2}{x}^{n - 6}</p><p>Nowgivensumis</p><p></p><p>Now given sum is</p><p>1 - C1 n \cdot 3 + C2 n \cdot {3}^{2} = 376</p><p></p><p>⇒ 1 - 3n + \dfrac{n\left(n - 1\right)}{2} \cdot 9 = 376</p><p></p><p>⇒ 1 - 3n + \dfrac{{n}^{2} - n}{2} \cdot 9 = 376</p><p></p><p>⇒ 2 - 6n + 9{n}^{2} - 9n = 752</p><p></p><p>⇒ 9{n}^{2} - 15n - 750 = 0</p><p></p><p>⇒ 3{n}^{2} - 5n - 250 = 0</p><p></p><p>⇒ n = \dfrac{5 \pm \sqrt{25 + 3000}}{6}</p><p></p><p>⇒ n = \dfrac{5 \pm 55}{6}</p><p></p><p>⇒ n = 10\left\{\text{ignoring negative sign}\right\}</p><p>Now,</p><p>Now,{T}_{r + 1} = {\left(- 1\right)}^{r} Cr 10 {3}^{r}{x}^{10 - 3r}</p><p>So,forcoefficientof</p><p>So, for coefficient of{x}^{4}wetake</p><p>we take </p><p>10 - 3r = 4</p><p></p><p>⇒ 3r = 6</p><p></p><p>⇒ r = 2</p><p>So,coefficientof</p><p>So, coefficient of{x}^{4}isgivenby,</p><p>is given by,</p><p>{T}_{2 + 1} = {\left(- 1\right)}^{2} \cdot C2 10 {3}^{2} = 45 \times 9 = 405.

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