If n=1 f(1)=2-7+2f(0)=-5+14=9.

f(2)=4-7+18=-3+18=15.

f(3)=6-7+30=29.

f(4)=8-7+58=59.

f(5)=10-7+118=121.

f(n)=2n-7+2f(n-1)=-7+2(n+f(n-1)). The factor of 2 operates with each successive evaluation of f(n), so 2ⁿ is going to be applied.

f(1)=2×1-7+2f(0)=2×1-7+2×7=2+7.

Since we are looking for a formula for n≥0, f(0) is the first term in the series. Since f(0)=7, we can expect 7 to be a constant term in the formula, and the remaining part of the formula to contain the factor n.

If we look at the series: 7, 9, 15, 29, 59, 121 and take the differences between consecutive pairs we get:

2, 6, 14, 30, 62

If we take half of these we get:

1, 3, 7, 15, 31 which can be represented by 2ⁿ-1. Therefore doubling gives 2(2ⁿ-1). And f(n)-f(n-1)=2(2ⁿ-1), so f(n)=f(n-1)+2(2ⁿ-1).

f(n-1)+2(2ⁿ-1)=2n-7+2f(n-1) from which

f(n-1)=2(2ⁿ-1)-2n+7, therefore f(n)=2(2ⁿ-1)-2n+7+2(2ⁿ-1)=4(2ⁿ-1)-2n+7.

Check formula for n=0 to 5:

f(0)=7, f(1)=9, f(2)=15, f(3)=29, f(4)=59, f(5)=121. The formula in the question doesn’t generate the correct series. So **f(n)=4(2ⁿ-1)-2n+7**.