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Binomial Coefficients

Find the rela...

Question

Find the relation between $r$ and $n$ in order that the coefficients of the $3 r$ th and $(r + 2)^{t h}$ terms of ${(1 + x)}^{2 n}$ may be equal.

r = n

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2 r = n

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r = - n

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r = - 4 n

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Solution

The correct option is B $2 r = n$
$⟹$ Coefficient of $(3 r)$ th tem
$=^{2 n} C_{3 r - 1}$
Coefficient of $(r + 2)$ th term
$^{2 n} C_{r + 1}$
$\frac{(2 n)!}{(3 r - 1)! (2 n - 3 r + 1)!} = \frac{(2 n)!}{(r + 1)! (2 n - r - 1)!}$
Now, by comparison:-
$(I)$ $2 n - 3 r + 1 = 2 n - r - 1$ or $3 r - 1 = r + 1$
$2 r = 2$
$r = 1$
$(I I)$ $r + 1 = 2 n - 3 r + 1$
$4 r = 2 n$
$r = \frac{n}{2}$
$∴ 2 r = n$

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Similar questions

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

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and coefficient of

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^{t h}

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are equal then value of

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Q. If the coefficients of

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Q. Given positive integers

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and the coefficients of

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are equal then

r =

Q. Given positive integers

r > 1, n > 2

and the coefficient of

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Find the relation between r and n in order that the coefficients of the 3rth and (r+2)th terms of (1+x)2n may be equal.

The correct option is B 2r=n⟹ Coefficient of (3r)th tem=2nC3r−1Coefficient of (r+2)th term2nCr+1(2n)!(3r−1)!(2n−3r+1)!=(2n)!(r+1)!(2n−r−1)!Now, by comparison:-(I) 2n−3r+1=2n−r−1 or 3r−1=r+12r=2r=1(II) r+1=2n−3r+14r=2nr=n2∴2r=n

Find the relation between $r$ and $n$ in order that the coefficients of the $3 r$ th and $(r + 2)^{t h}$ terms of ${(1 + x)}^{2 n}$ may be equal.