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Question

In the expansion of (1+ax)n, nN, then the coefficient of x and x2 are 8 and 24 respectively. Then?

A
a=2,n=4
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B
a=4,n=2
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C
a=2,n=6
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D
None of these
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Solution

The correct option is A a=2,n=4
Here in the expansion of (1+x.a)n,
nN, the (r+1)th term will be
Tr+1=(nr)anr.br
here first term a=1 &
second term b=x.a, so,
Tr+1=(nr)(1)nr.(x.a)r=(nr).xr.ar
But given that, coefficient of x is 8 so,
For x power 1, we have to put r=1
So, T2=(n1)a1.x1
So, (n1)a=8(i)
Now, coefficient of x2 is 24, so, r=2
T3=(n2)a2.x2
So, 24=(n2)a2(ii)
From (1) & (2),a.n=8 a=8/n
n(n1)2.a2=24
n(n1)2(8n)2=24
n2nn2×32=24
11n=2432=34, So, n=4 from (i)
a=2

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