Mathematical Induction questions (1 Viewer)

[pLuvia]

Prove he following statements by mathematical induction, where n is a natural number

1+2(2) + 3(2^2) + ..... n(2^n-1) = 1 +(n-1) 2^n

i got up to this answer

1 + 2^k (2k)

is that wrong?

 

[Trev]

1+2^k.2k = 1+2^k.2¹.k=1+k.2ⁿ⁺¹ which is the desired result when n=k+1. So no, it's not wrong...

 

[Pace_T]

Prove he following statements by mathematical induction, where n is a natural number

1+2(2) + 3(2^2) + ..... n(2^n-1) = 1 +(n-1) 2^n

i got up to this answer

1 + 2^k (2k)

is that wrong?

1 + 2^k (2k) = 1+ 2.k.2^k = 1 + k.2^(k+1)

now, in the original equaiton if n = k+ 1, RHS is 1 + k.2^(k+1)
which is the same, so its right i think

 

[pLuvia]

ok thanks guys

 

[pLuvia]

Where n is a natural number

41^n - 1 is divisible by 20

My working:

S(n) = 41^n - 1
When n = 1
S(1) = 40
Assume S(k) is divisible by 20
S(k) = 41^k - 1 = 20N [ Where N is an integer]
S(k+1) = 41^k+1 - 1
= 41(41^k) - 1
= 41(20N+1) - 1
= ???

What can I do now?

 

[KFunk]

= 41(20N+1) - 1
= ???

What can I do now?

= 41(20N+1) - 1
= 20(41N) + 41 - 1
= 20(41N) + 40
= 20(41N + 2)
= 20R, where R is an integer

 

[pLuvia]

Thank you

 

[pLuvia]

81(3²ⁿ) - 2²ⁿ is divisible by 5, where n is a natural number

 

[KFunk]

81(3²ⁿ) - 2²ⁿ is divisible by 5, where n is a natural number

Skipping past that n= 1 stuff... Let S_n be the proposition that 81(3²ⁿ) - 2²ⁿ is divisible by 5 i.e. that

81(3²ⁿ) - 2²ⁿ = 5M (where M is an integer)

Assume true for n = k

81(3^2k) - 2^2k = 5Q (multiply through by [4][9])

4.81(3^2k+2) - 9.(2^2k+2) = 5(36Q)

81(3^2k+2) - (2^2k+2) = 5(36Q) - (243.3^2k+2 - 8.2^2k+2)


RHS = 5(36Q) - (2187.3^2k - 32.2^2k)

= 5(36Q) - 27(81.3^2k - 2^2k) + 5.2^2k

= 5(36Q) - 5(27Q) + 5(2^2k)
= 5R , where R is an integer.

hence 81(3^2k+2) - (2^2k+2) = 5R

∴ S_k ==> S_k+1 etc...

 

[pLuvia]

81(32k) - 22k = 5Q (multiply through by [4][9])

4.81(32k+2) - 9.(22k+2) = 5(36Q)

81(32k+2) - (22k+2) = 5(36Q) - (243.32k+2 - 8.22k+2)

Why did you multiply it through by [4][9]

 

[KFunk]

Why did you multiply it through by [4][9]

To transform the S_k case into the S_k+1 case.

3^2k x 9 = 3^2(k+1)

2^2k x 4 = 2^2(k+1)

In the question you have 81(3^2k) - 2^2k and you have to make up the powers to 2(k+1) = 2k + 2 somehow so I multiplied through by 4x9. There's no point taking 9 out of 81 because then you have to make it 81 again by multiplying by nine .

 

[pLuvia]

oh okz, but can't you just make all the "k"s into "k+1" by substitution? That's the the way I learnt it

 

[KFunk]

Maybe... give it a go.

 

[pLuvia]

This is what I did,

Assuming S(k) is divisible by 5
S(k) = 81(3^2k) - 2^2k = 5N [ where N is an integer]
S(k+1)
= 81(3^2k+2) - 2^2k+2
= 81.3^2k.3² - 2^2k.2²
= 81.3^2k.9 - (81(3^2k) - 5N).4 [Sub 2^2k = 81(3^{2k - 5N})

Now I don't know what to do here?

 

[KFunk]

= 81.3^2k.9 - (81(3^2k) - 5N).4 [Sub 2^2k = 81(3^{2k - 5N})

Now I don't know what to do here?

= 81.3^2k.9 - (81(3^2k) - 5N).4

= 9.(81)(3^2k) - 4.(81)(3^2k) + 5(4N)

= 5.(81)(3^2k) + 5(4N)

= 5T ... where T is an integer.

The step that you're missing each time is just rearranging terms so that you can take out a factor of five. It was the same deal above for divisibility by 20.

 

[pLuvia]

Ok, thanks for pointing that out, I'll take that into account next time