IIT 79 Binomial Series

JEE Main Sample Questions › Algebra questions › IIT 79 Binomial Series

Show that C₁² - 2C₂² + 3C₃²............-2n.C_2n² = (-1)^n-1.nC_n

where C_r stands for ²ⁿC_r

#Mathematics #Algebra

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5 Answers

1

kunl ·2011-01-31 07:49:28

coefficeint of x^2n-1 in 2nx(1-x^2)^2n-1

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swordfish ·2011-01-31 09:33:39

How did you get that?

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21

Shubhodip ·2011-02-01 00:32:32

(1-x)²ⁿ = C₀ - C₁x + C₂ x² .....

differentiate w.r.t x

2n(1-x)^2n-1 = C₁ - 2C₂x + ...... "1"

again,

(1+x)²ⁿ =

C₀ x²ⁿ + C₁ x^2n-1 ............... "2"

multiplying 1 and 2 we get the required series which in turn is the co efficient on x^2n-1

in the expansion of 2n(1-x)^2n-1 (1+x)²ⁿ

;)

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swordfish ·2011-02-01 07:21:27

I tried but did not get how to find the coefficient of x^2n-1 in that expansion. Can you help?

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21

Shubhodip ·2011-02-02 04:45:23

co eff of x^2n-1 in 2n(1-x)^2n-1(1+x)^2n-1 (1+x )

= co eff of x^2n-1 in 2n(1-x²)^2n-1 (1+x )

= co eff of x^2n-1 in 2n(1-x²)^2n-1 + 2nx (1- x²)^2n-1

= co eff of x^2n-1 = 2nx(1- x²)^2n-1

= co eff of x^2n-2 in 2n(1-x²)^2n-1

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