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Algebra

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Joined:	7 Aug 2007
Post:	184

5 Apr 2008 10:08:00 IST

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579

one of the best questions of binomial...

None

(i.j) (ⁿC_i . ⁿC_j)

^0<i<j<=n

please post ur solutions..i dont know the answer....

(ⁿC_i ²+ ⁿC_j²)

^0<i<j<=n

i m getting its answer as (n+2)²ⁿC_n...so i wanted to confirm.....

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Comments (4)

anchit saini

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Joined: 1 Feb 2008

Posts: 1251

5 Apr 2008 10:25:41 IST

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edit
didn't read properly

GAURAV SHARMA

Hot goIITian

Joined: 7 Aug 2007

Posts: 184

5 Apr 2008 10:36:53 IST

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anchitsaini ...the answer given by u is valid if the question had been

(ⁿC_i . ⁿC_j)

note that here one more term is there i.j ....u didnt take that into account...and in the end u have made a mistake abt the sign...

GAURAV SHARMA

Hot goIITian

Joined: 7 Aug 2007

Posts: 184

5 Apr 2008 10:41:05 IST

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yaar lekin solve it now...

anchit saini

Blazing goIITian

Joined: 1 Feb 2008

Posts: 1251

5 Apr 2008 11:45:01 IST

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here is my shot at the answer -->

taking

(i.j) (ⁿC_i . ⁿC_j) = x

^0<i<j<=n

(1 + x)^n = (C_0 + C_1x + ................... + C_n x^n)

differentiating with respect to x

n(1 + x)

^(n-1)

^n-1 )

on putting x =1

n(2)

^n-1

or on squaring

n^2*2

^{2n - 2} =
(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 ) +

2[(1.2) C_1C_2 + (1.3)C_1C_3 + .........(1.n)C_1C_n

+ (2.3) C_2C_3 +.....................................................]

=(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 ) + 2x

now

we have to find the value of

(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 )

for that we proceed as follows --->

n(1 + x)

^(n-1)

^n-1 ) ---1

replacing x by 1/x

n(1 + 1/x)

^n-1=

^n-1 ) ---2

1 * 2

[n^2(1 + x)

^2n-2

^n-1

^n-1 ) *

^n-1 )

constant term in lhs is

n^2

* coefficient of x^n-1 in (1 + x)^(2n-2)]

= n^2

*^{2n - 2}

_n-1

constant term in rhs is -->

(C_1^2 + 2^2C_2^2 + ...........n ^2C_n^2 )

hence

^{2n - 2}

_n-1

thus

n^2

^{(2n - 2)} = n^2

^{2n - 2}

_n-1

^{(2n - 3)} - n^2

^{2n - 2}

_n-1

]

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