Biquadratic

using the given info...f(x) = (x-1)⁴

then (x+1)ⁿ = [λ(x)](x-1)⁴ + ax³ + bx² + cx + d (where λ(x) is a function of x though we dont need to bother abt this function)...

replacing x-1 = t...we get..

(t+2)ⁿ = ⁿC₀ tⁿ + ⁿC₁ t^n-1.2 +......+ ⁿC_n-3 t³.2^n-3 + ⁿC_n-2 t².2^n-2+ ⁿC_n-1 t.2^n-1 + ⁿC_n.2ⁿ

so now ⁿC_n-3 t³.2^n-3 + ⁿC_n-2 t².2^n-2+ ⁿC_n-1 t.2^n-1 + ⁿC_n 2ⁿ = at³ + (3a + b)t² + (3a + 2b + c)t + (a+ b+ c + d)

Now by comparing the L.H.S and R.H.S. we can get the value of a,b,c,d in terms of n... [1]

that seems rit

sorry for late reply.

We have

(x+1)^n = (x-1)^4 g(x) + A(x-1)^3 + B(x-1)^2 + C(x-1)+D

Put x=1 and we have D = 2^n

Differentiating once and putting x=1, we have

C= n2^{n-1}

In this manner we obtain B= n(n-1)2^{n-2}

and A= n(n-1)(n-3)2^{n-3}

Now to obtain a,b,c,d, we have a = A; b = -3A +B; c = 3A -2B+C; d = -A + B-C+D