Question

i hv been staring at these types of problems for a long time and im not really getting any of them rite...it would be nice if someone could help me.. thankx in advance..

Answer 1

5 + 11 + 17 +....+ (6n-1) = n(3n + 2)
1) Show true for n=1
6(1) - 1 = (1)(3(1)+2)
6 - 1 = 1 (3 +2)
5 = 5
2) Show true for n=k
5 + 11 +17 + ...+ 6n - 1 = n (3n + 2)
5 + 11 + 17 + ...+(6k-1) = k(3(k)+2)
3)Show true for n = k+1
5+11 +17 +...+(6k - 1)+ (6(k+1)-1) = (k+1)(3(k+1)+2)
k(3k+2) + 6(k) = (k+1)(3(k+1)+2)

I dont know where to go from here....help..plz :(

Answer 2

The directions are : use mathematical inductoin to prove that each statement is true for all natural numbers n

Answer 3

That's not quite how you do induction. Your step 1 is right, but steps 2 and 3 are off. You need to prove that it's true for n = k+1, assuming that it is true for n = k. So, you want to set n = k and then prove it for n = k+1.

Answer 4

Thats what i did..except i put "show" instead of "assume" :(

Answer 5

can you help me..?

Answer 6

Working on it, give me one minute.

Answer 7

okay, no problem

Answer 8

Oookay, I found the part that you messed up. You simplified (6(k+1)-1) to 6(k), but it should actually be 6k+5

Answer 9

Other than that you're on the right track.

Answer 10

\[ (6(k+1)-1) \to 6(k), \] simplify that!

Answer 11

So @Shaq if we fix that part, we'll have \(k(3k+2)+(6k+5)=(k+1)(3(k+1)+2)\). Next step is to simplify that equation.

Answer 12

omg thanku! i get it now; il ask if i need anyaditional help!

Answer 13

Right on, any time :)

Answer 14

so you always do the distribution before adding the additonal one?

Answer 15

Remember order of operations, PEMDAS, Parentheses Exponents Multiplication/Division Addition/Subtraction. Parentheses always come first :)

Answer 16

okay thanks! i think thats why i been getting these problems wrong. thanks a ton!

Answer 17

Sure thing :)