Question

0

Answer 1

you are trying to prove:
\[\sum_{i = 1}^{n} (-1)^{(n+1)} n^2 = 1^2 - 2^2 + 3^2 - .... + (-1)^{(n+1)} n^2 = (-1)^{(n+1)} \frac{ n(n+1) }{ 2 } \]

i would write it like this first

then establish your base case for i = 1
which is 1 = 1 so its true

do you understand what i did so far?

Answer 2

yes, i get confused on the induction proof of n+1

Answer 3

ok lets assume an arbitrary k where:

i messed up the notation before but this is the correct way to write it
\[\sum_{i = 1}^{k} (-1)^{(i+1)} i^2 = (-1)^{(k+1)} \frac{ k(k+1) }{ 2 } \]
is true

and prove for k+1
so you are trying to prove this now
\[\sum_{i = 1}^{k+1} (-1)^{(i+1)} i^2 = (-1)^{(k+1)} \frac{ (k+1)((k+1)+1) }{ 2 } \]

Answer 4

right

Answer 5

i messed up again sorry it should look like this

\[\sum_{i = 1}^{k+1} (-1)^{(i+1)} i^2 = (-1)^{((k+1)+1)} \frac{ k(k+1) }{ 2 } \]

Answer 6

why the additional +1?

Answer 7

ugh my bad

\[\sum_{i = 1}^{k+1} (-1)^{(i+1)} i^2 = (-1)^{((k+1)+1)} \frac{ (k+1)((k+1)+1) }{ 2 } \]

Answer 8

you want to show it works for any value of k, since you know the arbitrary k works then you need to prove that k+1 works as well so you need to replace all the k's with a k+1

Answer 9

so because it was already n+1 it should be n+1+1 or k

Answer 10

it should be just how i written it

Answer 11

Now you have to prove this:
\[\sum_{i = 1}^{k+1} (-1)^{(i+1)} i^2 = (-1)^{((k+1)+1)} \frac{ (k+1)((k+1)+1) }{ 2 } \]

look at the LHS:

\[\sum_{i = 1}^{k+1} (-1)^{(i+1)} i^2 = (-1)^{((k+1)+1)} \frac{ (k+1)((k+1)+1) }{ 2 }\]

\[\sum_{i = 1}^{k} (-1)^{(i+1)} i^2 + (-1)^{((k+1)+1)} (k+1)^2 = (-1)^{((k+1)+1)} \frac{ (k+1)((k+1)+1) }{ 2 }\]

\[(-1)^{(k+1)} \frac{ k(k+1) }{ 2 } + (-1)^{((k+1)+1)} (k+1)^2 = \] By inductive hypothesis

Answer 12

wait so you lost me, the last step you have there is it simplified and proven??

Answer 13

im not done, you need to make both sides look equal for it to fully proven

Answer 14

i got stuck somewhere and im trying to figure it out how to do it right now

Answer 15

ok np ty

Answer 16

this is what i had but i think its wrong
(-1)^(n+1)n(n+1)/2 + (-1)^(n+2)*(n+1)^2
(-1)^(n+1) [ n(n+1)/2 + (-1)(n+1)^2 ]
(-1)^(n+1) [ n(n+1)/2 - (n+1)^2 ]
(-1)^(n+1) [ (n(n+1) - 2(n+1)^2)/2 ]
(-1)^(n+1) [ (n^2 +n - 2n^2 -4n -2)/2 ]
(-1)^(n+1) [ (-n^2 -3n -2)/2 ]
(-1)^(n+1)(-1) [ (n^2 +3n +2)/2 ]
(-1)^(n+2) [ (n+1)(n+2)/2 ]

Answer 17

\[(-1)^{(k+1)} \frac{ k(k+1) }{ 2 } + (-1)^{(k+1)}(-1) (k+1)^2 = \]
\[(-1)^{(k+1)} (k+1) (\frac{ k }{ 2 } + (-1) (k+1)) =\] I factored out a \[(-1)^{(k+1)}(k+1)\]
\[(-1)^{(k+1)} (k+1) (\frac{ k }{ 2 } + \frac{ -2(k+1) }{ 2 }) = \]
\[(-1)^{(k+1)} (k+1) ( \frac{ -k-2 }{ 2 }) = \]
\[(-1)(-1)^{(k+1)} (k+1) ( \frac{ k+2 }{ 2 }) = \]
\[(-1)^{(k+2)} (k+1) ( \frac{ k+2 }{ 2 }) =\]
\[(-1)^{((k+1)+1)} (k+1) ( \frac{ ((k+1) + 1) }{ 2 }) =\]


There ya go, both sides are equal so this holds true for k+1
thus it hold true for n by induction

Answer 18

just follow understand what i did, its better to use summation for induction when doing these problems

Answer 19

alright thank you much! would you by chance want to look at another computer sci class problem i have :) ?

Answer 20

if its another math related i problem i think i can do it

Answer 21

and your welcome

Answer 22

want me to post it separate or here?

Answer 23

you can post it here

Answer 24

Define the sequence c1, c2... by the equations.
c1=c2=0 cn= c[n/3] +n for all n > 2 , prove the statement for all n>= 2 involving cn. the inductive step will assume truth of the statement c[n/3]
**thats basement c[n/3] not brackets**

Answer 25

can you define what a basement is?

Answer 26

basement or floor= greatest integer less than or equal to x, if [x] is basement

Answer 27

some refer to as floor

Answer 28

yea that what i thinking of

Answer 29

opposite is ceiling

Answer 30

so we apply the base case first
n = 2 so
\[c \lfloor \frac{ 2 }{ 3 } \rfloor + 2 = 2 \]
so its true

Answer 31

wait you are trying to prove all n > 2 right?
its cause you wrote >= 2 but 2 doesnt work actually since c0 is not defined yet

Answer 32

prove the statement for all n>= 2 involving cn. the inductive step will assume truth of the statement c[n/3] is what it says

Answer 33

alright im gonna reapply the base c for n = 3
c[3/3] + 3 = 3 true

Answer 34

is there anything else to the question?
i feel like something is missing

Answer 35

how is 2/3+2=2 true? or 3/3+3=3?

Answer 36

its not sorry i messed up

you are trying to prove:
c[n/3] + n = cn is true for all n >= 2

the thing is i dont know what c0 is equal to
because if we apply the base case to n = 2 then

c2 = c[2/3] + 2
0 = c0 + 2

Answer 37

it says c1=c2=0 sorry

Answer 38

cn= c[n.3]+n for all n>2, prove n>= n for involving cn.

Answer 39

what are the basis steps is the second part

Answer 40

im sorry i dont think i can help you with this, i cant seem to figure out the base case here

Answer 41

alright ty anyways

Answer 42

i would reccomend mathhelpforum.com and let the professionals try it hahaha

again sorry hope you can get the answer :(