Algebra - Binomial Expansion

Binomial Expansion

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Shaswata Roy ·Mar 21 '13 at 5:19

If we put x=1x we get,

$\frac{(x^{2}+x+1)^{n}}{x^{2n}}=\frac{a_{0}x^{2n}+a_{1}x+\dots +a_{2n}}{x^{2n}}$

However on expanding just the numerator we get that it is also equal to-

$\frac{a_{0}+a_{1}x+\dots +a_{n}x^{2n}}{x^{2n}}$

Therefore,
a₀ = a_2n
a₁ = a_2n-1

Or in general,
a_r = a_2n-r.......(i)

Given:
$(1+x+x^{2})^{n}=a_{0}+a_{1}x+\dots +a_{2n}x^{2n}\dots (ii)$

Put x= -x,
$(1-x+x^{2})^{n}=a_{0}-a_{1}x+\dots +a_{2n}x^{2n}\\=a_{0}x^{2n}-a_{1}x^{2n-1}\dots +a_{2n}\text{(from i)}\dots (iii)$

Put x = x²,

$(1+x^{2}+x^{4})^{n}=a_{0}+a_{1}x^{2}+\dots +a_{2n}x^{4n}\dots (iv)$

Coefficient of x²ⁿ in (iv) is = a_n

Multiplying (ii) and (iii),

$(1-x+x^{2})^{n}\times (1+x+x^{2})^{n}=(1+x^{2}+x^{4})^{n}\\(a_{0}+a_{1}x+\dots +a_{2n}x^{2n})\times (a_{0}x^{2n}-a_{1}x^{2n-1}\dots +a_{2n})$

Coeeficient of x²ⁿ in the RHS is,

$a_{0}^{2}-a_{1}^{2}\dots +a_{2n}^{2}$

We find from (iv) that this is equal to a_n

Again using the fact that a_r=a_n-r we may find,

$a_{0}^{2}-a_{1}^{2}+\dots +(-1)^{n-1}a_{n-1}=\frac{1}{2}(a_{0}^{2}+a_{2n}^{2}-a_{1}^{2}-a_{2n-1}\dots +(-1)a_{n}^{2})\\=\frac{1}{2}(a_{n}+(-1)^{n}a_{n}^{2})$

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Manish Shankar good work :)
Upvote·0· Reply ·Mar 22 '13 at 3:50
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