The first 3 terms in the expansion of (1+ax)n(n=0) are 1,6x and 16x2. Then the value of a and n are respectively

(1+ax)n=nCo1+nC1ax+nC2(ax)2∴nC1(ax)=6x∴nC2(ax)2=16x2∴(n−1)!n!×a=6∴xa=6∴(n−2)!2!n!a2=16∴(n)(n−1)a2=32∴(n−1)(a)1=163∴na−a1=163∴6−a=316∴a=36−16=32∴x=26×3=9∴x=9,a=32.

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The first 3 terms in the expansion of (1+ax)^{n}(n 0) are 1, 6

Question

The first $3$ terms in the expansion of $(1 + a x)^{n} (n \neq = 0)$ are $1, 6 x$ and $16 x^{2}$ . Then the value of $a$ and $n$ are respectively

2

and

9

3

and

2

2/3

and

9

3/2

and

6

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Michael Discussed

The first

3

terms in the expansion of

(1 + a x)^{n} (n \neq = 0)

are

1, 6 x

and

16 x^{2}

. Then the value of

a

and

n

are respectively

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Text solutionVerified

(1 + a x)^{n} =^{n} C_{o} 1 +^{n} C_{1} a x +^{n} C_{2} (a x)^{2}

∴^{n} C_{1} (a x) = 6 x

∴^{n} C_{2} (a x)^{2} = 16 x^{2}

∴ \frac{n !}{( n - 1 )!} \times a = 6

∴ x a = 6

∴ \frac{n ! a ^{2}}{( n - 2 )! 2 !} = 16

∴ (n) (n - 1) a^{2} = 32

∴ \frac{1}{( n - 1 ) ( a )} = \frac{3}{16}

∴ \frac{1}{na - a} = \frac{3}{16}

∴ 6 - a = \frac{16}{3}

∴ a = \frac{6 - 16}{3} = \frac{2}{3}

∴ x = \frac{6}{2} \times 3 = 9

∴ x = 9, a = \frac{2}{3}

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Question Text	The first $3$ terms in the expansion of $(1 + a x)^{n} (n \neq = 0)$ are $1, 6 x$ and $16 x^{2}$ . Then the value of $a$ and $n$ are respectively
Answer Type	Text solution:1
Upvotes	150

The first 3 terms in the expansion of (1+ax)n(n=0) are 1,6x and 16x2. Then the value of a and n are respectively

Text solutionVerified

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The first $3$ terms in the expansion of $(1 + a x)^{n} (n \neq = 0)$ are $1, 6 x$ and $16 x^{2}$ . Then the value of $a$ and $n$ are respectively