A sequence a1,a2,a3,…, is defined recursively by a1=1,a2=1, and for k≥3,ak=31ak−1+41ak−2.Evaluate a1+a2+a3+⋯.
Solution — click to reveal
Let S=a1+a2+a3+⋯. Then S=a1+a2+a3+a4+a5+⋯=1+1+(31a2+41a1)+(31a3+41a2)+(31a4+41a3)+⋯=2+31(a2+a3+a4+⋯)+41(a1+a2+a3+⋯)=2+31(S−1)+41S.Solving for S, we find S=4.