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Using the recursive formula, f(n+1)=2*f(n)+1, and given that f(1)=3, calculate the product of the digits in the number f(17).
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looks like a lot of work f(2) = f(1+1) = 2*f(1) +1 = 3 .. .. f(17) = ..
it's going to be lots of work
f(3) = 2*3+1 = 7 f(4) = 2*7+1 = 15 f(5) = 31 f(6) = 63 ..
looks like 2^n-1
I'm getting 262143. If you want to check your work.
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It's easy to note that. f(n)= 2^{n+1}-1
\[ f(n)= 2^{n+1}-1 \]
Or what he said.
Now I have memorized upto 2^{16} so here I have to do one small multiplication. but is there any way to do this in a less tedious way? what if f(2434343) was asked? and instead of product it was asked as sum?
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