For all positive integers n- show that 2 n C n -2 n C n -1-1-22 n -2 C n -1

L.H.S., = ^2nC_n + ^2nC_n 1 = 2n!/n! n! + 2n!/(n 1)! (2n n + 1)! = (2n)!/n(n 1)! n! + (2n)!/(n 1)! (n + 1)n! = (2n)!/n(n 1)! n! + (2n)!/(n 1)! (n + 1)n ...

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Byju's Answer

Standard XII

Mathematics

nCr Definitions and Properties

For all posit...

Question

For all positive integers n, show that

$^{2 n} C_{n} +^{2 n} C_{n - 1} = \frac{1}{2} (^{2 n + 2} C_{n + 1})$

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Solution

$L . H . S ., =^{2 n} C_{n} +^{2 n} C_{n - 1} = \frac{2 n!}{n! n!} + \frac{2 n!}{(n - 1)! (2 n - n + 1)!} = \frac{(2 n)!}{n (n - 1)! n!} + \frac{(2 n)!}{(n - 1)! (n + 1) n!} = \frac{(2 n)!}{n (n - 1)! n!} + \frac{(2 n)!}{(n - 1)! (n + 1) n!} = \frac{(2 n)!}{n (n - 1)! n!} + [\frac{1}{n} + \frac{1}{n + 1}] = \frac{(2 n)!}{n! (n - 1)!} [\frac{1}{n} + \frac{1}{n + 1}] = \frac{(2 n)!}{n! (n - 1)!} [\frac{2 n + 1}{n (n + 1)}] = \frac{(2 n + 1)!}{n! (n + 1)!} R H S = {\frac{1}{2}}^{2 n + 2} C_{n + 1} = \frac{1}{2} [\frac{(2 n + 2)!}{(n + 1)! (2 n + 2 - n - 1)!}] = \frac{1}{2} [\frac{(2 n + 2)!}{(n + 1)! (n + 1)!}]$
$= \frac{1}{2} [\frac{(2 n + 2)!}{(n + 1) n! (n + 1)!}] = \frac{1}{2} [\frac{2 (n + 1) (2 n + 1)!}{(n + 1) n! (n + 1)!}] = \frac{(2 n + 1)!}{n! (n + 1)!} ∴ L H S = R H S$

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Similar questions

Q. If

(1 + x)^{n} = C_{0} + C_{1} x + \dots . . + C_{n} x^{n}

, then the value of

\sum \sum 0 \leq r < s \leq n C_{r} C_{s}

is equal to

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} C_{n} =^{2 n} C_{n - 1}

^{2 n} C_{n + 1}

Q. If

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

then the value of

\sum \sum 0 \leq r < s \leq n {(C_{r} + C_{s})}_{r}^{2}

Q. Assertion :If 'n' is even then

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 1}

Reason:

^{2 n} C_{1} +^{2 n} C_{3} +^{2 n} C_{5} + . . . . . +^{2 n} C_{n - 1} = 2^{2 n - 2}

Q. Assertion :If

x =^{n} C_{n - 1} +^{n + 1} C_{n - 1} +^{n + 2} C_{n - 1} + . . . +^{2 n} C_{n - 1}

then

\frac{x + 1}{2 n + 1}

is integer Reason:

^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r} a n d^{n} C_{r}

is divisible by n if n and r are co-prime

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For all positive integers n, show that 2nCn+2nCn−1=12(2n+2Cn+1)

For all positive integers n, show that

$^{2 n} C_{n} +^{2 n} C_{n - 1} = \frac{1}{2} (^{2 n + 2} C_{n + 1})$