IIT JEE , AIEEE Forum - Mathematics , Algebra - A proof in Binomial Theorem

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Prove that

C₁²- 2C₂²+ 3C₃²- .......+.......+ 2n.C_2n²= (-1)^n-1n.C_n

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(1+x)²ⁿ = 1 + C₁.x + C₂.x² + C₃.x³ + ..........

Differentiate both sides wrt x :

2n(1+x)^2n-1 = C₁ + 2C₂.x + 3C₃.x² + .........

Now replace x by -x in the above equation :

2n(1-x)^2n-1 = C₁ - 2C₂.x + 3C₃.x² + ......... (1)

Now write (x+1)²ⁿ = C₀x²ⁿ + C₁x^2n-1 + C₂x^2n-2 + .......... (2)

Now multiply (1) and (2)

2n(1-x)^2n-1(x+1)²ⁿ = (C₁ - 2C₂.x + 3C₃.x² + ......).(C₀x²ⁿ + C₁x^2n-1 + C₂x^2n-2 + ......)

Now expand RHS, you'll find that given expression is coefficient of x^2n-1 in expansion of RHS.

It should also be equal to coefficient of x^2n-1 in LHS.

LHS = 2n(1-x)^2n-1(x+1)²ⁿ = 2n(1-x²)^2n-1(1+x)

= 2n(1-x²)^2n-1 + 2nx(1-x²)^2n-1

1st term contains even powers of x and hence doesn't contain x^2n-1

coefficient of x^2n-1 in 2nd term = (-1)^n-1(2n)(^2n-1C_n-1)

so given expression = (-1)^n-1(2n)(^2n-1C_n-1) =

= (-1)^n-1(2n).{(2n-1)! / (n-1)!.n!}

Write (2n-1)! = (2n)! / 2n and (n-1)! = n!/n and put them in the abpve expression.

given expression = (-1)^n-1(2n).{(2n)! / 2(n!.n!)}

= (-1)^n-1n(²ⁿC_n)

Objective Approach :

Put n=1 in expression and options given and match them. If you get the answer its well and good. If you are getting two or three options correct try to match for n=2.

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