Aditya Gupta
Last Activity: 6 Years ago
(1+x+x2)n = a0 +a1x +a2x2 + a3x3+ .......+ a2nx2n... (1)
replace x by – 1/x
(1 – 1/x+1/x2)^n = a0 – a1/x +a2/x2 – a3/x3+ .......+ a2n/x2n
or (1 – x+x2)n = a0x^2n – a1x^(2n – 1) +a2x(2n – 2) – .......+ a2n... (2)
multiply 1 and 2
((1+x+x2)(1 – x+x2))^n= (a0 +a1x +a2x2 + a3x3+ .......+ a2nx2n)(a0x^2n – a1x^(2n – 1) +a2x(2n – 2) – .......+ a2n)
or (1+x^2+x^4)^n= (a0 +a1x +a2x2 + a3x3+ .......+ a2nx2n)(a0x^2n – a1x^(2n – 1) +a2x(2n – 2) – .......+ a2n)... (3)
now in 1, if we replace x by x^2, then we get
(1+x^2+x^4)^n= a0 +a1x^2 +a2x^4 + a3x^6+ .......+ a2nx^4n
put this value in 3,
a0 +a1x^2 +a2x^4 + a3x^6+ .......+ a2nx^4n= (a0 +a1x +a2x2 + a3x3+ .......+ a2nx2n)(a0x^2n – a1x^(2n – 1) +a2x(2n – 2) – .......+ a2n)
now consider the coefficient of x^2n on LHS. it is clearly an.
now consider the coefficient of x^2n on RHS. it is a0*a0+a1*( – a1)+a2*a2 – …..+a2n*a2n or a0^2-a1^2+a2^2-a3^2+...+a2n^2. since LHS=RHS, the coeffs on both sides should also equal each other.
hence a0^2-a1^2+a2^2-a3^2+...+a2n^2 = an