Help i dont understan Mathematical Induction thing someon help PLEASE?

Question

PROVE:

3 + 7 + 11 +.... +(4n - 1) = n(4n - 1)

Answer 1

n=1
LHS = 3
RHS = 1*(4-1)=3
LHS=RHS

n=2
LHS = 3 + 7 = 10
RHS = 2(8-1) = 14
LHS ≠ RHS

The RHS should have been n(2n+1)

3 + 7 + 11 + ...(4n-1) = n(2n+1)
Check if it is true
n=1
LHS = 3
RHS = 1*(2+1) = 3

n = 2
LHS = 7 + 3 = 10
RHS = 2*(4+1) = 10

Assume it is true for n=m
3 + 7 + 11 + ... (4m-1) = m(2m+1)
The next term would be 4(m+1) - 1 = 4m+3
Add 4m+3 to both sides
3 + 7 + 11 + .... [4(m+1) - 1] = m(2m+1) + 4m+3

RHS
m(2m+1) + 4m+3
= 2m^2 + m + 4m + 3
= 2m^2 + 5m + 3
= 2m+2 + 2m + 3m + 3
= 2m(m+1) + 3(m+1)
= (m+1)(2m+3)
= (m+1)(2(m+1) + 1)

If LHS = RHS for n=m, then LHS = RHS for n=m+1
LHS = RHS for n=1
Therefore it must be true for n=2
if it is true for n=2 then it must be true for n=3
.
.
.
It is true for all values of n

Answer 2

Question should read as n.(2n + 1) on RHS

P(n):-
3 + 7 + 11 + ---(4n - 1) = n.(2n + 1)
Assume true for k, a value of n

P(k):-
3 + 7 + 11 + ---(4k - 1) = k.(2k + 1)
Must now show P(1) and P(k + 1) to be true.

Consider P(1)
LHS = 3
RHS = 1 x 3 = 3
Thus P(1) is true

Consider P(k + 1)
3 + 7 + 11 ----+ (4k + 3) = (k + 1).(2k + 3)

Must now show this to be true.
3 + 7 + 11 -----+ (4k - 1) = k.(2k + 1)
Add next term to both sides:-
3+7 + (4k - 1)+(4k + 3) = k.(2k + 1) + (4k + 3)
3 + 7----(4k - 1) + (4k + 3) = 2k² + k + 4k + 3
3 + 7 ---(4k - 1) + (4k + 3) = 2k² + 5k + 3
3 + 7 ---(4k - 1) + (4k + 3) = (2k + 3).(k + 1)
ie have shown P(k + 1) is true

Thus
P(1) is true
P(k) is true
P(k + 1) is true
Therefore P(n) is true.

(Thank goodness for that having spent ages trying to prove an incorrect question. Now going to lie down in a darkened room for a short while!)

Answer 3

The formula should state that the sum of the series given to n terms is n(2n + 1).

Demonstrate that this formula works for n = 1.
The sum to 1 term is 3, and the value of n(2n + 1) with n = 1 is also 3.

Now assume that the formula is true for n = k, and prove that it is true for n = k + 1.
3 + 7 + 11 + .... + (4k - 1) = k(2k + 1)
Add the next term 4(k + 1) - 1 to each side of the equation.
3 + 7 + 11 + ......+ 4(k + 1) - 1 = k(2k + 1) + 4(k + 1) - 1
You now need to demonstrate that
k(2k + 1) + 4(k + 1) - 1 ....(1)
is the same as k(2k + 1) with k increased by 1, or that it is equal to:
(k + 1)(2(k + 1) + 1) ............(2)

Simplifying (1) gives:
2k^2 + 5k + 3 ..........(3)
Simplifying (2) gives:
2(k + 1)^2 + k + 1
= 2k^2 + 4k + 2 + k + 1
= 2k^2 + 5k + 3 ........(4)

As (3) and (4) match, the formula has been proved.

Answer 4

The equation for the series is -:

(4n - 1)

i.e. when n=1, 4 - 1 = 3
when n = 2, 4*2 - 1 = 7
when n = 3, 4*3 - 1 = 11
etc....etc.....
and this fits your series.

Answer 5

Let the statement be 3 + 7 + 11 + ...+ (4n-1) = n(4n-1).
When n=1, RHS = 1(4-1)
= 3
= LHS
Let us assume statement is true for n = k; i.e.
3 + 7 + 11 + ... + (4k-1) = k(4k-1)
To prove statement is also true for n = k + 1, i.e.
3 + 7 + 11+ ... + (4k-1) + (4k+3) = (k+1)(4k+3)
LHS = k(4k-1) + (4k+3)
= 4k^2 - k + 4k + 3
= 4k^2 + 3k + 3
But RHS = (k+1)(4k+3)
= 4k^2 +7k +3
There is something wrong indeed ... ... .

Answer 6

Sorry, your equation ∑[i=1,n] (4i – 1) = n(4n – 1) is a fallacy.
The correct equation is ∑[i=1,n] (4i – 1) = n(2n + 1).

Notational Convention: ∑[i=1,n] (4i – 1) is to be read as the sum of 4i - 1 from i = 1 to n while ∑[i=1,n-1] (4i – 1) is to be read as the sum of 4i - 1 from i = 1 to n-1.


Proof by mathematical induction PMI):

Case i = 1:

4(1) – 1 = 1[4(1) – 1]
4 – 1= 1[4 – 1]
3= 3

Case i > 1:

We have

∑[i=1,n] (4i – 1) = ∑[i=1,n-1] (4i – 1) + 4n – 1

Substitute ∑[i=1,n] (4i – 1) = n(2n + 1) to get

n(2n + 1) = ∑[i=1,n-1] (4i – 1) + 4n – 1

Finally, rearrange to get

∑[i=1,n-1] (4i – 1)

= n(2n + 1) – 4n + 1
= 2n2 + n – 4n + 1
= 2n2 – 3n + 1
= (n – 1)(2n – 1)
= (n – 1)(2n – 2 + 1)
= (n – 1)[2(n-1) + 1]

End-of-Proof

Answer 7

I think there's something wrong with your equation.

If you let n = 2
3+7 = 11

while n(4n-1) = 2[4(2)-1]=14 .

Check your problem again.

Answer 8

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Answer 9

The second answer is right.

Basically what you are doing is proving the equation is true for 1 particular case: n=1

Then assume true for n=m

Then prove true for n=m+1

If it is true for n=m+1 and you know it is tue for one particular value (in this case we chose 1) then you know it is true for all values!

Some nice logic there I think you'll agree.

Answer 10

i think your problem is defective