Let an=1+(1+n1)2+1+(1−n1)2.Compute a11+a21+a31+⋯+a1001.
Solution — click to reveal
We have that an1=1+(1+n1)2+1+(1−n1)21=(1+(1+n1)2+1+(1−n1)2)(1+(1+n1)2−1+(1−n1)2)1+(1+n1)2−1+(1−n1)2=1+(1+n1)2−1−(1−n1)21+(1+n1)2−1+(1−n1)2=n41+(1+n1)2−1+(1−n1)2=4n(1+(1+n1)2−1+(1−n1)2)=4n2+(n+1)2−n2+(n−1)2,so an1=4n2+(n+1)2−(n−1)2+n2.Hence, a11+a21+a31+⋯+a1001=412+22−02+12+422+32−12+22+432+42−22+32+⋯+41002+1012−992+1002=420201−1.