solve that

• Nov 22nd 2008, 04:12 AM
hoger
solve that
1) show that 9/ (n^3+(n+1)^3+(n+2)^3) ,n belongs N ,throhg
a) induction b)direct

2)show that

57/(7^n+2 +8^2n+1) , n blongs N , throgh

a) induction b)congrunce
• Nov 22nd 2008, 10:29 AM
Soroban
Hello, hoger!

Here's the first one . . .

Quote:

1) Show that $\bigg[n^3+(n+1)^3+(n+2)^3\bigg]$ is a multiple of 9, for $n \in N$

a) by induction

$S(n)\!:\;\;n^3 + (n+1)^3 + (n+2)^3$ is a multiple of 9.

Verify $S(1)\!:\;\;1^3+2^3+3^3 \:=\:36$, which is a multiple of 9.

Assume $S(k)\!:\;\;k^3+(k+1)^3 + (k+2)^3 \:=\:9a$, for some integer $a.$

Add $(k+3)^3 - k^3$ to both sides:

. . $k^3 +(k+1)^3 + (k+2)^3 + {\color{blue}(k+3)^3 - k^3} \;=\;9a + {\color{blue}(k+3)^3 - k^3}$

. . $\underbrace{(k+1)^3 + (k+2)^3 + (k+3)^3}_{\text{the left side of }S(k+1)} \;=\;9a + (k+3)^3 - k^3$

We must show that the right side is a multiple of 9.

The right side is: . $9a + k^3 + 9k^2 + 27k + 27 - k^3$

. . $= \;9a + 9k^2 + 27k + 27 \;=\;\underbrace{9(a + k^2 + 3k + 3)}_{\text{a multiple of 9}}$

The inductive proof is complete.

Quote:

b) by direct proof.
$N \;=\;n^3 + (n+1)^3 + (n+2)^3 \;=\;n^3 + (n^3 + 3n^2 + 3n + 1) + (n^3 + 6n^2 + 12n + 8)$

. . $= \;3n^3 + 9n^2 + 15n + 9 \;=\;3(n^3 + 3n^2 + 5n + 3) \;=\;3(n+1)(n^2+2n+3)$

We must show that either $(n+1)$ or $(n^2+2n+3)$ is a multiple of 3.

There are three cases for $n$

. . $\begin{array}{ccc}(1) & n\text{ is one less than a multiple of 3:} & n \:=\:3a - 1\\
(2) &n\text{ is a multiple of 3:} & n\:=\:3a \\
(3) &n\text{ is one more than a multiple of 3:} & n \:=\:3a+1 \end{array}$

$(1)\;\;n \:=\:3a-1$
. . .Then: . $n + 1 \:=\:(3a-1) + 1 \:=\:3a$, a multiple of 3.

$(2)\;\;n \:=\:3n$
. . . $\text{Then: }\;n^2+2n+3 \;=\;(3a)^2 + 2(3a) + 3 \:=\:9a^2 +6a + 3 \:=\: \underbrace{3(3a^2+2a+1)}_{\text{multiple of 3}}$

$(3)\;\;n \:=\:3a+1$
. . . $\text{Then: }\;n^2 \;+\; 2n\;+\;3 \;=\;(3a+1)^2 + 2(3a+1)\;+\; 3 \;=\; 9a^2 + 12a \;+\; 6 \;=\;\underbrace{3(3a^2 + 4a+2)}_{\text{multiple of 3}}$

. . $Q.E.D.$