First of all, f(1)=2. Then, further values of f are given by f(n)=f(n−1)+n.
Under these circumstances, which of these formulas gives f(n)?
(a) n+1, (b) (n+1)2, (c) 2n2+n+2, (d) 2(n+1)2, (e) 2(3n−1)2.
Solution
Of course we can just plug the first few integers into the five given formulas and see which works, but here is a more subtle way.
We start by calculating some values:
f(1)f(2)f(3)f(4)f(5)=====2,2+24+37+411+5====4,7,11,16.
With each step we are adding one more number so f should be something like f(n)≈1+2+3…+n. Let's calculate this sum:
1,1+21+2+31+2+3+41+2+3+4+5====3,6,10,15.
So, we see that
f(n)=(1+2+…+n)+1=2n(n+1)+1=2n2+n+2.
The correct answer is (c). (That formula for the sum of the first n whole numbers is a handy one to know.)
For inquiring minds: what if f(1) isn't 2? What if f(1) is, say, 3? or f(1)=n?