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Given the integers r>1,n>2 and coeff of (3r)^{th} and (r+2)^{n

Question

Given the integers $r > 1, n > 2$ and coeff of $(3 r)^{t h}$ and $(r + 2)^{n d}$ terms in the binomial expansion of $(1 + x)^{2 n}$ are equal, then

n = 2 r

n = 4 r

n = 2 r + 1

None of these

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Eleanor Discussed

Given the integers

r > 1, n > 2

and coeff of

(3 r)^{t h}

and

(r + 2)^{n d}

terms in the binomial expansion of

(1 + x)^{2 n}

are equal, then

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Text solutionVerified

Solution:

We know that in the coefficient of

r^{t h}

term in expansion of

(1 + x)^{n} =^{n} C_{r - 1}

∴

coefficients of

(3 r)^{t h}

and

(r + 2)^{t h}

in the expansion

(1 + x)^{2 n}

are

^{2 n} C_{3 r - 1}

and

^{2 n} C_{r + 1} .

Now,

^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1}

or,

3 r - 1 = r + 1

or,

3 r - 1 + r + 1 = n

(∵^{n} C_{r} =^{n} C s

then

r = s

r + s = n)

On solving,

3 r - 1 = r + 1

or,

r = 1

and

3 r - 1 + r + 1 = n

or,

4 r = n

or,

n = 4 r

Hence, D is the correct option.

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Question Text	Given the integers $r > 1, n > 2$ and coeff of $(3 r)^{t h}$ and $(r + 2)^{n d}$ terms in the binomial expansion of $(1 + x)^{2 n}$ are equal, then
Answer Type	Text solution:1
Upvotes	150

Given the integers r>1,n>2 and coeff of (3r)th and (r+2)nd terms in the binomial expansion of (1+x)2n are equal, then

Text solutionVerified

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Given the integers $r > 1, n > 2$ and coeff of $(3 r)^{t h}$ and $(r + 2)^{n d}$ terms in the binomial expansion of $(1 + x)^{2 n}$ are equal, then