First slide

Binomial theorem for positive integral Index

Question

The sum of rational terms in the expansion of ${(\sqrt{2} + 3^{1 / 5})}^{10}$ is

Moderate

A

250
B

41
C

530
D

none of these

Solution

General term in the given expansion of ${(\sqrt{2} + 3^{1 / 5})}^{10}$ is

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

$T_{r + 1} =^{10} C_{r} {(\sqrt{2})}^{10 - r} {(3^{1 / 5})}^{r}$

$T_{r + 1} =^{10} C_{r} {(2)}^{\frac{10 - r}{2}} {(3)}^{^{\frac{r}{5}}}, r = 0, 1, 2, ..., 10$

This will be a rational number if booth $\frac{10 - r}{2}$ and $\frac{r}{5}$ are whole numbers

$\frac{10 - r}{2} = 0 \Rightarrow r = 10$

$\frac{r}{5} = 0 \Rightarrow r = 0$

i.e. if $r = 0$ and $r = 10$

So, there are only two rational terms, namely

$T_{0 + 1}$ and $T_{10 + 1}$

Sum of these terms $= T_{1} + T_{11}$

$=^{10} C_{0} 2^{5} +^{10} C_{10} 3^{2}$ $(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$

$= 32 + 9 = 41$

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