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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2015 · Mathematics · Binomial Theorem

In the expansion of ( x + 1 3 x^2 )^ 10 the value of constant term...

In the expansion of (x+13x2)10\left(\sqrt{\mathrm{x}}+\frac{1}{3 \mathrm{x}^2}\right)^{10} the value of constant term (independent of x ) is

Options

  1. A5
  2. B8
  3. C45
  4. D90

Correct answer

A. 5

Step-by-step solution

Let rth \mathrm{r}^{\text{th }} term is independent of x . Tr=nCrxryn−r=10Cr(x)r(13x2)10−r=10Cr(13)10−r⋅(x)r(1x2)10−r\begin{aligned} & T_r={ }^n C_r x^r y^{n-r} \\ & ={ }^{10} C_r(\sqrt{x})^r\left(\frac{1}{3 x^2}\right)^{10-r} \\ & ={ }^{10} C_r\left(\frac{1}{3}\right)^{10-r} \cdot(\sqrt{x})^r\left(\frac{1}{x^2}\right)^{10-r} \end{aligned} Equating the coefficient of xx to zero. ⇒xr/2⋅x−2(10−r)=x0⇒r2−20+2r=0⇒52r=20⇒r=8\begin{aligned} & \Rightarrow \quad x^{\mathrm{r} / 2} \cdot \mathrm{x}^{-2(10-\mathrm{r})}=\mathrm{x}^0 \\ & \Rightarrow \quad \frac{\mathrm{r}}{2}-20+2 \mathrm{r}=0 \\ & \Rightarrow \quad \frac{5}{2} \mathrm{r}=20 \Rightarrow \mathrm{r}=8 \end{aligned} Coefficient =10Cr(13)10−r=10C8(13)10−8\begin{aligned} \text{ Coefficient } & ={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{3}\right)^{10-\mathrm{r}} \\ & ={ }^{10} \mathrm{C}_8\left(\frac{1}{3}\right)^{10-8} \end{aligned}=\frac{10 \times 9}{2} \times \frac{1}{9}=5$
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