Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.
1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n (6n^2 - 3n - 1)/2

Answer 1

hey do you have an idea how to start this ??

Answer 2

`1st)` substitute n for 1 to check is the L.H.S = R.H.S ? if the statement is true then next step is
`2nd)` assume it is true for n=k (substitute n for k) this step is called " induction assumption'
`3rd)` substitute n for k+1 and we want to show the statement is true for n= k+1 based on the 2nd step assumption

Answer 3

just tag me when u rready

Answer 4

@Nnesha

Answer 5

I substituted n with 1 on the left hand side to get \[\frac{ 1[6(1)^2-3(1)-1] }{ 2 }\]

Answer 6

Which would then make the LHS = 1

Answer 7

I dont know what to do with the RHS

Answer 8

I mean... I did that to the RHS and dont know what to do with the LHS

Answer 9

for first step
(3n - 2)^2 = n (6n^2 - 3n - 1)/2
you need to see if both sides are equal when u substitute n for 1

Answer 10

\[\large\rm (3(1)-2)^2=\frac{ 1[6(1)^2-3(1)-1] }{ 2 }\] both sides are equa l? then move to the next step

Answer 11

substitute n for k???

Answer 12

yes right that would be the 2nd step
and have to assume that n = k is true

Answer 13

What is k?

Answer 14

k is just a variable that represent the highest term
kth term

Answer 15

not the highest i mean one of the term

Answer 16

So...\[(3k-2)^2=\frac{ 1[6(k)^2-3(k)-1]}{ 2 }\]

Answer 17

\[(3k−2)2=k[6(k)2−3(k)−1]2\] *

Answer 18

oops...... \[(3k−2)^2=\frac{k[6(k)^2−3(k)−1] }{ 2 }\]

Answer 19

\[\rm 1^2 + 4^2 + 7^2 + ... + (3\color{red}{k} - 2)^2 = \frac{\color{Red}{k} (6\color{Red}{k}^2 - 3\color{Red}{n} - 1)}{2}\]
for 2nd step keep the first terms bec this is the assumption we need for 3rd step

Answer 20

we assumed that n =k is true
now move on to the 3rd step substitute all n's with k+1

Answer 21

is the 3n supposed to be 3n or 3k?

Answer 22

oh ye iforgot to change that

Answer 23

\[1^2+4^2+7^2+...+(3(k+1)−2)^2=\frac{ (k+1)[6(k+1)^2−3(k+1)−1] }{ 2 }\]

This is starting to look a little bit less pleasant

Answer 24

\[\rm 1^2 + 4^2 + 7^2 + ... + (3\color{red}{k} - 2)^2 = \frac{\color{Red}{k} (6\color{Red}{k}^2 - 3\color{Red}{k} - 1)}{2}\]
now substitute n for k+1
and we should leave the last term which is (3k-2)^2 at left side
\[\rm 1^2 + 4^2 + 7^2 + ... + \color{blue}{(3\color{blue}{k} - 2)^2}+(3[\color{red}{k+1}]-2)^2 = \frac{\color{Red}{k+1} (6\color{Red}{[k+1]}^2 - 3\color{Red}{[k+1]} - 1)}{2}\]

Answer 25

hahah ik but it will be fun!!

Answer 26

that's correct but we need to leave (3k-2)^2 at left side and add (3[k+1]-2)^2 because here we gotta use the assumption

Answer 27

\[\rm \color{orange }{1^2 + 4^2 + 7^2 + ... + (3\color{orange}{k} - 2)^2} = \frac{\color{Red}{k} (6\color{Red}{k}^2 - 3\color{Red}{k} - 1)}{2}\]
according to the 2nd assumption (the orange side ) is equal to
k(6k^2+3k-1)/2
so can substitute (the orange part) with k(6k^2+3k-1)/2

\[\rm \color{orange}{\frac{k(6k^2+3k-1)}{2}}+(3[\color{red}{k+1}]-2)^2 = \frac{\color{Red}{k+1} (6\color{Red}{[k+1]}^2 - 3\color{Red}{[k+1]} - 1)}{2}\]

Answer 28

make sense ? if no feel free to ask question :)

Answer 29

Attempting to solve. Dont think I know how.

Answer 30

well did you understand the last part ???

Answer 31

yes

Answer 32

I understood the substitution part

Answer 33

alright

Answer 34

remember the PEMDAS rule
parentheses first
then exponent
try to solve :=))

Answer 35

LHS= \[\frac{ 6k^3+3k^2-k }{ 2 }+(3[k+1]-2)^2\]

Answer 36

good so far :=))

Answer 37

\[(3[k+1]−2)^2\] becomes \[9k^2-1\]?

Answer 38

\[\frac{ 6k^3+3k^2−k }{ 2 } + 9k^2 -1\]

Answer 39

Not sure about the other side

Answer 40

I'll try to finish this tomorrow...

Answer 41

ugh sorry i was afk let me chec k

Answer 42

hmmm well it can't be negative one
it's not difference of squares \[a^2-b^2=(a-b)(a+b)\]

Answer 43

\[(a+b)^2=a^2+2ab+b^2\]
first distribute bracket by 3

Answer 44

alright i'll be free tomorrw :=))

Answer 45

@Nnesha Do you know if it is true or false?

Answer 46

The world may never know...