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KEAM 2011 Math Paper

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Question : 100 of 120

Marks: +1, -0

If

(1+ax)^n=1+6x+\frac{27}{2}x^2+\dots+a^n x^n,

then the values of

a

n

are respectively

$2,3$
$3,2$
$\frac{3}{2},4$
$1,6$
$\frac{3}{2},6$

Solution:

Given,

(1+a x)^{n}=1+6 x+\frac{27}{2} x^{2}+\dots +a^{n} x^{n}

....(i)

The expansion of

(1+a x)^{n}

is,

(1+a x)^{n}=1+n a x+\frac{n(n-1)}{2!}(a x)^{2}+\dots

......(ii)

On comparing the coefficient of like powers of

x

in Eqs. (i) and (ii),

n a = 6

.....(iii)

\frac{27}{2} = \frac{n(n-1)}{2} \cdot a^{2}

\Rightarrow 27 = (n-1)(n a) \cdot a

27 = (n-1) a 6

[from Eq. (iii)]

(n-1) a = \frac{9}{2}

.....(iv)

From Eqs. (iii) and (iv),

\frac{(n-1)6}{n} = \frac{9}{2}

\Rightarrow \frac{n-1}{n} = \frac{3}{4}

\Rightarrow 4 n - 4 = 3 n

\Rightarrow n = 4

From Eq. (iii),

a = \frac{6}{4}

\Rightarrow a = \frac{3}{2}

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