Question

Use mathematical induction to prove the statement is true for all positive integers n.

The integer n^3 + 2n is divisible by 3 for every positive integer n.

Answer 1

By induction. Just say since it happens for n consecutive numbers, it's happens for all other numbers.

Answer 2

1. for n=1 statement is true, we can check it: 1+2=3 => 3 divides 3
2. Assume that the statement is true for n=k: 3 divides k^3+2k
3. Then it has to be right for n=k+1 - let's check
(k+1)^3+2(k+1)=k^3+3k^2+5k+3=(k^3+2k) + (3k^2+3k+3) = (k^3+2k) + 3(k^2+k+1) - we can divide this by 3, so the statement is proved

Answer 3

done thank yoU!!

Answer 4

What happened to my medal?

Answer 5

No problem :)

Answer 6

mathematical induction, you so sexy

Answer 7

@blurbendy Ahem!

Answer 8

hehe

Answer 9

Let P(n)=n^3+2n
for n=1,1^{3}+2*1=3 which is divisible by 3
Hence P(1) is true.
Assume that P(k) is true.
P(k)=k^3+2k is divisible by 3
k^3+2k=3l where l is an integer.
k^3=3l-2k
P(k+1)=(k+1)^3+2(k+1)
\[=k ^{3}+1^{3}+3k*1\left( k+1 \right)+2k+2\]
\[=k ^{3}+1+3k ^{2}+3k+2k+2\]
\[=k ^{3}+2k+3k ^{2}+3k+3=3l-2k+2k+3\left( k ^{2} +k+1\right)\]
\[=3\left( l+k ^{2}+k+1 \right)=3*an integer.\]
Hence P(k+1) is true .
Hence by induction P(n)is true forall n

Answer 10

i like that response better lol