Find the first integral term in the expansion of -3-29

The correct option is C Let the first integral term be T_r+1 T_r+1 = ^9C_r (3^1/2)^9 r (2^1/3)^r = (3^9 r/2) (2^r/3) We get the first integral term when r=3(B ...

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Standard XI

Mathematics

Number of Terms in Binomial Expansion

Find the firs...

Question

Find the first integral term in the expansion of $(\sqrt{3} + \sqrt[3]{2})^{9}$

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Solution

The correct option is D

Let the first integral term be $T_{r + 1}$
$T_{r + 1} =^{9} C_{r} (3^{\frac{1}{2}})^{9 - r} (2^{\frac{1}{3}})^{r}$
= $(3^{\frac{9 - r}{2}}) (2^{\frac{r}{3}})$
We get the first integral term when r=3(Both $\frac{r}{3} a n d \frac{9 - r}{2}$ should be integers. One way of proceeding is start with 0, then 1,2,3. When r=3, we get the first integral term)
$\Rightarrow T_{4} =^{9} C_{4} 3^{3} 2^{1}$
$= 54^{9} C_{4}$

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Find the first integral term in the expansion of (√3+3√2)9

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