Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

1) 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6

2) 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 1

i'll do the first one. and u can do the second one. OK?

Answer 2

Okay I'll try :)

Answer 3

let's try the second.
it seems you have some typos in the first one.
check that out.!!

Answer 4

we have to prove for all n natural that
\[ \large 1^2+4^2+7^2+\dots+(3n-2)^2=\frac{n(6n^2-3n-1)}{2} \]

Answer 5

(i) n=1 (induction base). If n=1 the left side is
\[ \large 1^2=1 \]
and the right side is
\[ \large \frac{1(6\cdot1^2-3\cdot1-1)}{2}=\frac{6-3-1}{2}=\frac{2}{2}=1 \]
they are equal, hence we have our induction base.
OK?

Answer 6

Okay...

Answer 7

Now, let n>1 and suppose that (induction hypothesis)
\[ \large 1^2+4^2+7^2+\dots+(3n-2)^2=\frac{n(6n^2-3n-1)}{2} \]
we have to prove that (induction thesis)
\[ \large 1^2+\dots+(3n-2)^2+[3(n+1)-2]^2=\frac{(n+1)[6(n+1)^2-3(n+1)-1]}{2} \]
OK?

Answer 8

oh sorry the right side should be
\[ \large =\frac{(n+1)[6(n+1)^2-3(n+1)-1]}{2} \]

Answer 9

let's reduce a little bit:
\[ \large =\frac{(n+1)[6(n^2+2n+1)-3n-3-1]}{2} \]
\[ \large =\frac{(n+1)(6n^2+9n+2)}{2} \]
so our induction thesis is
\[ \large 1^2+\dots+(3n-2)^2+(3n+1)^2=
\frac{(n+1)(6n^2+9n+2)}{2} \]

r u following so far??

Answer 10

@aleisha0301 ??

Answer 11

Oh yes I am! Sorry

Answer 12

Ok. let's probe the thesis:
\[ \large 1^2+\dots+(3n-2)^2+(3n+1)^2= \]
\[ \large =\frac{n(6n^2-3n-1)}{2}+(3n+1)^2 \]
\[ \large =\frac{n(6n^2-3n-1)+2(9n^2+6n+1)}{2} \]
\[ \large =\frac{6n^3-3n^2-n+18n^2+12n+2}{2} \]
\[ \large =\frac{6n^3+15n^2+11n+2}{2} \]

ok so far?

Answer 13

Yes I understand so far :)

Answer 14

we r gonna use synthetic division to factor the last numerator into what we r supposed to get:
\[ \large \begin{array}{r|rrrr}
& 6 & 15 & 11 & 2\\
-1 & & -6 & -9 & -2 \\
\hline
& 6 & 9 & 2 & \fbox{0}
\end{array} \]
this gives us
\[ \large =\frac{6n^3+15n^2+11n+2}{2}=
\frac{(n+1)(6n^2+9n+2)}{2} \]
which is precisely what we were supposed to get.
QED.

Answer 15

@aleisha0301 ??

did u follow all of this??

Answer 16

I did. So the last part is the answer?

Answer 17

everything is the answer. it is a demonstration by induction.

Answer 18

Hm okay. Thank you for taking time to help me! :)

Answer 19

glad to help.

Answer 20

now for the first one, there is something wrong.

u have
\[ \large 4\cdot6+5\cdot7+6\cdot8+\dots+4n(4n+2)=
\frac{4(4n+1)(8n+7)}{6} \]

Answer 21

do u see the last summand on the left side.??

Answer 22

namely 4n(4n+2)

Answer 23

@aleisha0301 ??

Answer 24

In the problem it's 4(4n+1)(8n+7)/6

Answer 25

no, on the left side. ignore the right side.

Answer 26

Ohh! it's 4(4n+2). My mistake!

Answer 27

why don't repost the problem (properly written) and will work on it.
OK??

Answer 28

Okay :)

Answer 29

just "call" me.

Answer 30

I want to say thank you this help me understand so much

Answer 31

so the statement is true for all positive integers? @helder_edwin

Answer 32

@aleisha0301

Answer 33

I don't know, @Ally_B . She never posted the problem with the corrections.