Yes, Kaya.. A positive integer n has an odd number of divisors if and only if n is a … 4k + 1. ) . 1. 2 n−1. ∑ k=0. (. 4n. 4(n – 1 – k)1. ) … xn = nxn−1 + n n – 2.
Hope that helped!
LOL
.
Nice to see a new face in the strip, though, it would have been cool seeing Fay, again.
Yes, Kaya.. A positive integer n has an odd number of divisors if and only if n is a … 4k + 1. ) . 1. 2 n−1. ∑ k=0. (. 4n. 4(n – 1 – k)1. ) … xn = nxn−1 + n n – 2.
Hope that helped!
LOL.
Nice to see a new face in the strip, though, it would have been cool seeing Fay, again.