The term independent of x in the expansion of 2 x-1-x10A- 6520B- 8464- 6720D- 8064

The correct option is D 8064Let (r+1)th term be the term free from x So, T_r+1=^10C_r(2x)^10 r(1/x)^r =^10C_r(2)^10 r(x)^10 2r nbsp; Since, this term is free ...

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Standard XII

Mathematics

Sum of Product of Binomial Coefficients

The term inde...

Question

The term independent of $x$ in the expansion of ${(2 x + \frac{1}{x})}^{10}$ is

6720

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8064

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Solution

The correct option is B $8064$
Let $(r + 1)$ th term be the term free from $x$
So,
$T_{r + 1} =^{10} C_{r} (2 x)^{10 - r} {(\frac{1}{x})}^{r} =^{10} C_{r} (2)^{10 - r} {(x)}^{10 - 2 r}$
Since, this term is free from $x$
$∴ 10 - 2 r = 0 \Rightarrow r = 5$
$(5 + 1)$ th term free is from $x$
$\Rightarrow T_{6} =^{10} C_{5} 2^{5} = 8064$

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The term independent of x in the expansion of (2x+1x)10 is

The correct option is B 8064Let (r+1)th term be the term free from x So, Tr+1=10Cr(2x)10−r(1x)r=10Cr(2)10−r(x)10−2r Since, this term is free from x ∴10−2r=0⇒r=5 (5+1)th term free is from x ⇒T6=10C525=8064

The term independent of $x$ in the expansion of ${(2 x + \frac{1}{x})}^{10}$ is