Compute the sum Σ=1+12⋅328⋅12−1+1+32⋅528⋅22−1+⋯+1+20052⋅200728⋅10032−1
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For v=1,2,…,1003, we have 1+(2v−1)2(2v+1)28v2−1=(4v2−1)2(4v2−1)2+8v2−1=(4v2−1)216v4=(4v2−14v2)2 Hence we can write 1+(2v−1)2(2v+1)28v2−1Σ=(4v2−14v2)2=4v2−14v2=4v2−14v2−1+1=1+(2v−1)(2v+1)1=1+2(2v−1)(2v+1)2v+1−(2v−1)=1+21(2v−11−2v+11).=1+21(1−31)+1+21(31−51)+⋯+1+21(20051−20071)=1003+21(1−20071)=1003+20071003=20071003⋅2008=20072014024.