If f(1) = 1 and f(n+1) = 2f(n) + 1 if n≥1, then f(n) is equal to 2^n+1b) 2^n c) 2^n-1d)2^(n-1)-1

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`f(1)=1 and f(n)=2sum_(r=1)^(n-1) f (r)`. Then `sum_(n=1)^mf(n)` is equal to `f(1)=1 and f(n)=2sum_(r=1)^(n-1) f (r)`. Then `sum_(n=1)^mf(n)` is equal to