Falls "Induktion" tatsächlich gar nicht in der Tag-Liste stehen sollte, dann geht es so :
∑ [k=1 .. n] k·5k = f(5) für f(x) = ∑ [k=1 .. n] k·xk
f(x) = ∑ [k=1 .. n] k·xk = ∑ [k=0 .. n] k·xk
= x·∑ [k=0 .. n] k·xk-1 = x·∑ [k=0 .. n] (xk)'
= x·(∑ [k=0 .. n] xk)' = x·((xn+1-1)/(x-1))'
= x·((n+1)·xn·(x-1) - (xn+1-1)) / (x-1)2
f(5) = 5·((n+1)·5n·4 - 5n+1 + 1) / 42
= ( (n+1)·4·5n+1 - 5·5n+1 + 5 ) / 16
= ( 5n+1·(4n+4 - 5) ) / 16 + 5/16