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umka2103 [35]
3 years ago
8

Use mathematical induction to prove the statement is true for all positive integers n. The integer n3 + 2n is divisible by 3 for

every positive integer n.
Mathematics
2 answers:
Tema [17]3 years ago
5 0
1. prove it is true for n=1
2. assume n=k
3. prove that n=k+1 is true as well


so

1.
\frac{n^3+2n}{3}=
\frac{1^3+2(1)}{3}=
\frac{1+2}{3}=1
we got a whole number, true


2.
\frac{k^3+2k}{3}
if everything clears, then it is divisble


3.
\frac{(k+1)^3+2(k+1)}{3} =
\frac{(k+1)^3+2(k+1)}{3} =
\frac{k^3+3k^2+3k+1+2k+2)}{3}=
\frac{k^3+3k^2+5k+3)}{3}
we know that if z is divisble by 3, then z+3 is divisble b 3
also, 3k/3=a whole number when k= a whole number

\frac{k^3+2k}{3} + \frac{3k^2+3k+3}{3}=
\frac{k^3+2k}{3} + k^2+k+1=
since the k²+k+1 part cleared, it is divisble by 3

we found that it simplified back to \frac{k^3+2k}{3}

done



Troyanec [42]3 years ago
4 0

Answer:

We have to use the 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.

  • for n=1

n^3+2n=1+2=3 is divisible by 3.

Hence, the statement holds true for n=1.

  • Let us assume that the statement holds true for n=k.

i.e. k^3+2k is divisible by 3.---------(2)

  • Now we will prove that the statement is true for n=k+1.

i.e. (k+1)^3+2(k+1) is divisible by 3.

We know that:

(k+1)^3=k^3+1+3k^2+3k

and 2(k+1)=2k+2

Hence,

(k+1)^3+2(k+1)=k^3+1+3k^2+3k+2k+2\\\\(k+1)^3+2(k+1)=(k^3+2k)+3k^2+3k+3=(k^3+2k)+3(k^2+k+1)

As we know that:

(k^3+2k) was divisible as by using the second statement.

Also:

3(k^2+k+1) is divisible by 3.

Hence, the addition:

(k^3+2k)+3(k^2+k+1) is divisible by 3.

Hence, the statement holds true for n=k+1.

Hence by the mathematical induction it is proved that:

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

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