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in Binomial Theorem by (50.8k points)
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Show that the expansion of (x2 + 1/x)12 does not contain any term involving x-1.

1 Answer

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Best answer

Given as

(x2 + 1/x)12

If x-1 occurs at the (r + 1)th term in the given expression.
Now, we have:

Tr+1 = nCr xn-r ar

Now, for this term to contain x-1, we must have

24 – 3r = -1

3r = 24 + 1

3r = 25

r = 25/3

It is not possible, since r is not an integer.

Thus, there is no term with x-1 in the given expansion.

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