解决T（n-1）+ sqrt（n）的递归

Question

I'm hoping that I'm going about this problem the correct way. It asks to solve the recurrence:

T(n) = T(n-1) + sqrt(n)

So far I have researched and been able to get to this point:

T(n) = T(n-2) + (n-1) + sqrt(n) T(n) = T(n-3) + (n-2) + (n-1) + sqrt(n) T(n) = T(0) + 1 + 2 + ... + (n-2) + (n-1) + sqrt(n)

I'm having trouble understanding what the pattern may be to solve for 1+2+...+sqrt(n)

Answer 1

You start with unrolling the recursion and you should receive a sum of square roots. The sum of square roots is a generalized harmonic number and yours one can be approximated with:

Answer 2

The second line is already wrong.

If T (n) = T (n - 1) + sqrt (n), then T (n - 1) = T (n - 2) + sqrt (n - 1), therefore

T (n) = T (n - 2) + sqrt (n - 1) + sqrt (n)

T (n) = T (n - 3) + sqrt (n - 2) + sqrt (n - 1) + sqrt (n)

T (n) = T (n - 4) + sqrt (n - 3) + sqrt (n - 2) + sqrt (n - 1) + sqrt (n)

and so on.

The sum of the square roots from 1 to n is about the same as the integral of sqrt (x) from 1 to n.