@5SpeedDeasil to [email protected] • 1 year agoRandom internet people explaining math better then math teacheri.imgur.comimagemessage-square198arrow-up11.74Karrow-down162cross-posted to: math
arrow-up11.68Karrow-down1imageRandom internet people explaining math better then math teacheri.imgur.com@5SpeedDeasil to [email protected] • 1 year agomessage-square198cross-posted to: math
minus-square@only_lurkinglink1•1 year agoFor the case that n = 0 (before the first run of the loop), x(0) = 1. For the first actual case, n = 1. X(1) = x(0)*3*n = 1*3*1 = 3. For the next case, n = 2. X(2) = x(1)*3*n = 3*3*2 = 18. For the next case, n = 3. X(3) = x(2)*3*n = 18*3*3 = 162. For the next and last case, n = 4. X(4) = 162*3*4 which I’m not computing. The computer value of x(4) is the value of the product loop. If that doesn’t help, I could try helping again to rephrase, but I’m not sure what else to add.
For the case that n = 0 (before the first run of the loop), x(0) = 1.
For the first actual case, n = 1. X(1) = x(0)*3*n = 1*3*1 = 3.
For the next case, n = 2. X(2) = x(1)*3*n = 3*3*2 = 18.
For the next case, n = 3. X(3) = x(2)*3*n = 18*3*3 = 162.
For the next and last case, n = 4. X(4) = 162*3*4 which I’m not computing. The computer value of x(4) is the value of the product loop.
If that doesn’t help, I could try helping again to rephrase, but I’m not sure what else to add.