proving method of differences

Question

Please help me solve this method of differences question

Answer 1

If we substitute n+1 for n in n²-n+1 we get: (n²+2n+1)-(n+1)+1=n²+n+1, which is the denominator of the second term.

Let a_n=1/(n²-n+1), b_n=1/(n²+n+1), then a_n+1=b_n.

u_n=a_n-b_n; u_n+1=a_n+1-b_n+1=b_n-b_n+1; u_n+2=a_n+2-b_n+1=b_n+1-b_n+2; etc.

u_n+u_n+1+u_n+2=a_n-b_n+b_n-b_n+1+b_n+1-b_n+2=a_n-b_n+2, because intermediate terms cancel out.

Note that S₁=u₁=a₁-b₁ and S₂=u₁+u₂=a₁-b₁+b₁-b₂=a₁-b₂.

Also S_n=a₁-b₁+b₁-b₂+...-b_n-1+b_n-1-b_n=a₁-b_n.

When n=m+1, u_m+1=a_m+1-b_m+1=b_m-b_m+1; when n=2m, u_2m=a_2m-b_2m=b_2m-1-b_2m.

The number of terms between the limits m+1 (inclusive) and 2m=2m-(m+1)+1=m, which defines the sum for S_m.

All that remains now is to calculate and simplify the difference between the first and last terms of the summed series.

First term is a_m+1=1/(m²+2m+1-m-1+1)=1/(m²+m+1); last term is b_2m=1/(4m²+2m+1).

Therefore S_m=1/(m²+m+1)-1/(4m²+2m+1),

S_m=(4m²+2m+1-m²-m-1)/[(m²+m+1)(4m²+2m+1)],

S_m=(3m²+m))/[(m²+m+1)(4m²+2m+1)]=m(3m+1))/[(m²+m+1)(4m²+2m+1)].