Question

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

Answer 1

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

Answer 2

I think its false because...

4(4(1)+1)(8(1)+7))/6=
4(5)(15)/6=50

Answer 3

Pease help @Nnesha

Answer 4

Please help @dan815

Answer 5

if we rewrite your identity with n=1, we get:
left side= 4n(4n+2)=4*1(4*1+2)=4*6=24

Answer 6

please wait a moment

Answer 7

I thought that when you did n=1, you were suppose to compare it to the first initial term

Answer 8

we have to rewrite the left side setting n=1

Answer 9

whereas the right side, with n=1, is:
\[\frac{{4\left( {4n + 1} \right)\left( {8n + 1} \right)}}{6} = \frac{{4 \times 5 \times 9}}{6} = 24\]

Answer 10

Ok what do I do with the 24?

Answer 11

since with n=1, the left side is equal to the right side, then our identity is true for n=1

Answer 12

How is the left side= right side??

Answer 13

left side =24
and right side =24
when n=1

Answer 14

Ohh I think you made a typo because its supposed to be (8n +7) @Michele_Laino

Answer 15

it is true, sorry

Answer 16

then, the right side is:
\[\frac{{4\left( {4n + 1} \right)\left( {8n + 7} \right)}}{6} = \frac{{4 \times 5 \times 15}}{6} = 50\]

Answer 17

so, since the left side is 24
whereas the right side is 50
then we can conclude that your identity is false

Answer 18

Ok makes perfect sense thank you!

Answer 19

thank you! :)

Answer 20

Do you mind helping me with another problem @Michele_Laino ?

Answer 21

yes!

Answer 22

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

Answer 23

here we have to apply the same procedure of your previous exercise. So I set n=1, and I write your left side:

Answer 24

Ok (3(1)- 2)^2= 1^2

Answer 25

that's right!
\[{\left( {3n - 2} \right)^2} = {\left( {3 - 2} \right)^2} = {1^2} = 1\]

Answer 26

now I write your right side for n=1:

Answer 27

\[\frac{{n\left( {6{n^2} - 3n - 1} \right)}}{2} = \frac{{1 \times \left( {6 - 3 - 1} \right)}}{2} = \frac{{1 \times 2}}{2} = 1\]

Answer 28

Ok so 1(6(1)^2-3(1)-1)/2= 6-3-1/2 = 2/2=1 yes got same answer

Answer 29

ok! So your identity is true for n=1. Now let's suppose your identity true for a generic n

Answer 30

ok that's where I do n=k. right?

Answer 31

yes!

Answer 32

Now I write the left side for n+1, namely:
\[\begin{gathered}
{1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3\left( {n + 1} \right) - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 3 - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 1} \right)^2} \hfill \\
\end{gathered} \]

Answer 33

\[\Large \begin{gathered}
{1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3\left( {n + 1} \right) - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 3 - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 1} \right)^2} \hfill \\
\end{gathered} \]

Answer 34

Ok so for the right side it is (3n-2)^2 +(3n+1)^2 = n+1 (6(n+1)^2-3(n+1)-1)/2

Answer 35

that's right!
we have:
\[\Large \begin{gathered}
{1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3\left( {n + 1} \right) - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 3 - 2} \right)^2} = \hfill \\
= {1^2} + {4^2} + {7^2} + ... + {\left( {3n - 2} \right)^2} + {\left( {3n + 1} \right)^2} = \hfill \\
\hfill \\
= \frac{{n\left( {6{n^2} - 3n - 1} \right)}}{2} + {\left( {3n + 1} \right)^2} = \hfill \\
\end{gathered} \]

Answer 36

Ok this is the part where I get confused as to what's next

Answer 37

Do I have to simplify further?

Answer 38

\[\begin{gathered}
\frac{{n\left( {6{n^2} - 3n - 1} \right)}}{2} + {\left( {3n + 1} \right)^2} = \hfill \\
= \frac{{6{n^3} + 15{n^2} + 11n + 2}}{2} = \frac{{\left( {n + 1} \right)\left( {6{{\left( {n + 1} \right)}^2} - 3\left( {n + 1} \right) - 1} \right)}}{2} \hfill \\
\end{gathered} \]

Answer 39

now, we can note that the last expression, is the right side, of your original identity, with n+1 in place of n

Answer 40

Oh ok now I see! So I have to prove that the left and right sides are equal. Right?

Answer 41

yes! we can state that your original identity is true for all \[n \in \mathbb{N}\]

Answer 42

What do the symbols above mean?

Answer 43

\[n \in \mathbb{N}\] means for n which belong to the set of all natural number, or the set of whole number. That set is indicated, with this symbol:
\[\Large \mathbb{N}\]

Answer 44

belongs*

Answer 45

Ok so all of the steps above proved the statement true?

Answer 46

yes! generally, the standard procedure, is to verify that your identity for n=2 and for n=3 also

Answer 47

oh wait can you help me simplify n+1 (6(n+1)^2-3(n+1)-1)/2....im not getting it to equal the left side

Answer 48

ok!

Answer 49

we can write:
\[\large \begin{gathered}
\frac{{\left( {n + 1} \right)\left( {6{{\left( {n + 1} \right)}^2} - 3\left( {n + 1} \right) - 1} \right)}}{2} = \hfill \\
\hfill \\
\frac{{\left( {n + 1} \right)\left( {6\left( {{n^2} + 2n + 1} \right) - 3\left( {n + 1} \right) - 1} \right)}}{2} = \hfill \\
\hfill \\
= \frac{{\left( {n + 1} \right)\left( {6{n^2} + 12n + 6 - 3n - 3 - 1} \right)}}{2} = \hfill \\
\hfill \\
= \frac{{\left( {n + 1} \right)\left( {6{n^2} + 9n + 2} \right)}}{2} = \hfill \\
\hfill \\
= \frac{{6{n^3} + 9{n^2} + 2n + 6{n^2} + 9n + 2}}{2} = \hfill \\
\hfill \\
= \frac{{6{n^3} + 15{n^2} + 11n + 2}}{2} = \hfill \\
\end{gathered} \]

Answer 50

Wow I was distributing it in a weird way to where I got 6k^ 2 +12k+6.....Now I see what you did

Answer 51

Thank you so much @Michele_Laino for helping me solve the problems step by step!

Answer 52

Thank you! :) @lisa123