Is it possible to find $f(n)$?

Question

Is it possible to find $f(n)$?

$$\sum_{n=1}^{k}f(n)=2×3^k$$

I dont know, how to begin.

Answer 1

$$\sum_{n=1}^{k}f(n)=2×3^k$$ Implies that for $k>1$, $$f(k) = \sum_{n=1}^{k}f(n) - \sum_{n=1}^{k-1}f(n) = 2×3^k - 2×3^{k-1} = 4×3^{k-1}$$ and $f(1) = 6$

Answer 2

You jsut need to rewrite your formula... $$\sum_{n=1}^{k}f(n)=2×3^k$$ $$\sum_{n=1}^{k-1}f(n)+\sum_{n=k}^{k}f(n)=2×3^k$$ $$\sum_{n=1}^{k-1}f(n)+f(k)=2×3^k$$ $$f(k)=2×3^k-\sum_{n=1}^{k-1}f(n)$$ $$f(k)=2×3^k-2\times 3 ^{k-1}$$ Or if you wanna use n $$f(n)=2×3^n-2\times 3 ^{n-1}=4\times 3 ^{n-1}$$

For n>1