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Question

Find the term independent of x in the expansion of (x2+2x4)(x+1x)32


A

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B

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C

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D

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Solution

The correct option is A


(x2+2x4)(x+1x)32 can be written as (x2)(x+1x)32+(2x4)(x+1x)32

To find the term independent of x, we find the terms independent of x in (x2)(x+1x)32 and (2x4)(x+1x)32 and add them.

Consider (x2)(x+1x)32, the term independent of x in this expansion will have a power of x equal to -2 in the expansion of (x+1x)32 (Then only the power of x in (x2)(x+1x)32 will become zero, because there is an x2 term outside)

Tr+1 = 32Crx32r(1x)r

= 32Crx322r

We want the power of x to be -2

32 - 2r = -2

34 = 2r

r = 17

The term independent of x in the expansion of (x2)(x+1x)32 is 32C17.................(1)

Similarly for (2x4)(x+1x)32, we have to find a term which has the power of x equal to -4 in (x+1x)32 to get a term independent of x in the expansion (2x4)(x+1x)32.

Tr+1 = 32Crx32r(1x)r

= 32Crx322r

We want the power of x to be -4

32-2r = -4

2r = 36

r = 18

The term independent of x in the expansion of (2x4)(x+1x)32 is 232C18...............(2)

Combining (1) and (2), we get,

32C17+232C18


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