Question

If the sum of the binomial coefficients of the expansion (2x+1/x)^n is equal to 256 then the term independent of x is

Answer 1

Answer:

Step-by-step explanation:

the sum of the binomial coefficient of (a+b)n is 2n .

here 2n=256=28⇒n=8

now let (r+1)th term in the expansion of [2x+1x]8 is independent of x.

Tr+1=Cr8.(2x)8−r.(1x)r=Cr8.28−r.x8−r−r=Cr8.28−r.x8−2r

this is constant term , therefore exponent of x must be zero.

8−2r=0⇒2r=8i.e. r=4

then constant term = C48.28−4=8*7*6*54*3*2*24

=70*16=1120

Answer 2

If the sum of the binomial coefficients of the expansion $(2x+\frac{1}{x})^{n}$ is equal to 256, then the term independent of x is 1120.

Step-by-step Explanation:

Given:

The sum of the binomial coefficients of the expansion $(2x+\frac{1}{x})^{n}$ is 256.

To be found:

To find the independent term of x from the binomial expansion of the given expression.

Solution:

We know that the sum of the binomial coefficients is equal to $2^{n}$

So, $2^{n} =256$

$\implies 2^{n} = 2^{8}\\ \implies n=8$

The total number of terms in the given binomial expansion $=n+1=8+1=9$

Also, we know that the general term of binomial expression is given as  $T_{r}= ^{n}C_{r}(a)^{n-r}(b)^{r}$  --------(1)

Substituting the given values in (1), we get

$T_{r}= ^{8}C_{r}(2x)^{8-r}(\frac{1}{x} )^{r}$

Rearranging the terms, we get

$T_{r+1}= ^{8}C_{r}(2)^{8-r} (x)^{8-r}(x)^{-r}$

$\implies T_{r+1}= ^{8}C_{r}(2)^{8-r} (x)^{8-2r}$  -------(2)

Here, we have to find the term that is independent of x.

$(x)^{8-2r}=x^{0} \\\implies 8-2r=0$

Evaluating, we get

$2r=8\\\implies r=4$

Now, substituting the values in (2), we get

$T_{4+1}= ^{8}C_{4}(2)^{8-4} (x)^{8-8}$

$\implies T_{5}= \frac{8\times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} (2)^{4} (x)^{0}\\\implies T_{5}=\frac{26880}{24} \\\implies T_{5}= 1120$

Therefore, the term independent of x in the given binomial expansion is 1120.

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