First let us rewrite it as
∑k≥13k+22k+1=322∑k≥13k2k=92∑k≥1(32)k
The last series is a geometric series, for which we have
∑k≥1qk=1−qq (which we can deduce from ∑k=1nqk=1−qq−qn+1 and qn+1→0 for ∣q∣<1) and therefore the result is given by
92∑k≥1(32)k=92×1−3232=94
Comments
Leave a comment