hello
on the jpg attachment...how does
2^(k-1)+2^(k-2)+2^(k-3) becomes 2^0+2^1+…+2^(k-3)+2^(k-2)+2^(k-1) ?
thank you
they just added $\displaystyle 2^0 + 2^1 + \cdots$ for the sake of using the sum of the geometric series. it is just manipulation, a trick we can use to our advantage. don't let this jar you too much, the proof is valid. in fact, a similar trick was used earlier in the proof, obviously you didn't have a problem with it then. when proving P(2), they claimed: $\displaystyle a_2 = 3 \le {\color{red}1} + 3 = 4 = 2^2$. where did that 1 come from? they just added it. does that make sense? of course, surely if we are less than 3 we would be less than 3 + 1 = 4. it is the same trick used here. $\displaystyle 2^0 + 2^1 + \cdots$ will be a positive number. so surely, if we are less than $\displaystyle 2^{k - 1} + 2^{k - 2} + 2^{k - 3}$ then we will be less than $\displaystyle \underbrace{2^0 + 2^1 + \cdots}_{\mbox{a positive number}} + 2^{k - 1} + 2^{k - 2} + 2^{k - 3}$