If the constant term- in binomial expansion 2xr-1x2 10 is 180- then r is equal to

T_k+1= ^10C_k.(2x^r)^10 k. ( 1x^2 )^k= ^10C_k.2^10 k.x^(10 k)r 2k ∴ nbsp; (10 k)r 2k=0 and ^10C_k.2^10 k=180 ⇒ ^10C_k.2^10 k=45.2^2 ⇒ k=8 and r = 8 ...

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Byju's Answer

Standard XII

Mathematics

Finding Integral Part of Numbers of the Form a^b Where a Is Irrational

If the consta...

Question

If the constant term, in binomial expansion ${(2 x^{r} + \frac{1}{x^{2}})}^{10}$ is $180$ , then $r$ is equal to

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Solution

$T_{k + 1} =^{10} C_{k} . (2 x^{r})^{10 - k} . {(\frac{1}{x^{2}})}^{k} =^{10} C_{k} {.2}^{10 - k} . x^{(10 - k) r - 2 k}$
$∴ (10 - k) r - 2 k = 0 and^{10} C_{k} {.2}^{10 - k} = 180$
$\Rightarrow^{10} C_{k} {.2}^{10 - k} = {45.2}^{2}$
$\Rightarrow k = 8$ and $r = 8$

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If the constant term, in binomial expansion (2xr+1x2)10 is 180, then r is equal to

Tk+1= 10Ck.(2xr)10−k.(1x2)k= 10Ck.210−k.x(10−k)r−2k ∴(10−k)r−2k=0 and 10Ck.210−k=180 ⇒ 10Ck.210−k=45.22 ⇒ k=8 and r=8

If the constant term, in binomial expansion ${(2 x^{r} + \frac{1}{x^{2}})}^{10}$ is $180$ , then $r$ is equal to

$T_{k + 1} =^{10} C_{k} . (2 x^{r})^{10 - k} . {(\frac{1}{x^{2}})}^{k} =^{10} C_{k} {.2}^{10 - k} . x^{(10 - k) r - 2 k}$
$∴ (10 - k) r - 2 k = 0 and^{10} C_{k} {.2}^{10 - k} = 180$
$\Rightarrow^{10} C_{k} {.2}^{10 - k} = {45.2}^{2}$
$\Rightarrow k = 8$ and $r = 8$