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Question

The number of terms in the expansion of (x2+1+1x2)n, nϵN, is

A
2n
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B
3n
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C
2n+1
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D
3n+1
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Solution

The correct option is A 2n+1
(x2+1+1x2)n
=(x2+2+1x21)n
=[(x+1x)21]n
=[(x+1x+1)(x+1x1)]n
=(x+1x+1)n((x+1x1)n
Hence there will be 2n+1 terms.
Alternatively, considering (x2+1+1x2)n
If we replace the 1 by 2, we will get 2n terms dependent on x, since one term will be independent of x. However, since x2+1+1x2 is not a perfect square, there won't be any term independent of x except 1 itself. Hence there will 2n+1 terms.

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