Just Observe it -2 !!!!!

We will show, by induction, that $f(3n)=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ . The case $n=1$ is given. Now $f(3n)=n+f\left(3(n-1)\right)=n+\sum_{k=1}^{n-1}k=\sum_{k=1}^{n}k$ as claimed.

Thus $f(300)=\frac{100\times101}{2}=5050$ and the sum of the digits is $\boxed{10}$ .

make those into coordinate points to find an equation for it using a graphing calculator (1,1) (2,3) (3,6) (4,10) ... (term number, term value) let t stand for term value and n stand for term number the equation is [t=0.5n^2+0.5n] to figure out the 100th term just plug in 100 for n, the term number t=0.5(100)^2+0.5(100) t=5050

If you program at all, you look at this and see it as a recursive algorithm quickly. Basically, it has (n = 3 --> 1) as the base case, and goes up by 3 from the base case until it reaches the starting input, adding (x/3) each time.

Pulling out all those pesky threes will show you the true algorithm is basically "f(x) = x + f(x - 1), starting at f(100)".

Do a quick Gauss Summation to get f(100) = 5050, and add 5 and 5 for a final answer of 10.