Induction questions help (1 Viewer)

[jellybelly59]

9.Write a formula for the sum of the first n multiples of 4 and use the method of induction to prove your formula true.

10. The formula for the sum of n terms of an arithmetic series Sn= n/2 [ 2a + (n-1)d] where a is the first term and d the common difference. Prove this formula true by mathematical induction

11. Use mathematical induction to prove that for any positive integer n, (5^n-1) is divisible by 4

 

[jellybelly59]

bump

 

[chumbumm. =D]

lols!
your soo funny.

Location: where there is pho and sugar cane drinks

srry. i noe this doesnt answer ur question..

 

[shantu1992]

11. Use mathematical induction to prove that for any positive integer n, (5^n-1) is divisible by 4

that question is wrong

 

[Trebla]

Q9 is the lame question...
4 + 8 + 12 + 16 + ..... + 4n = 4(1 + 2 + 3 + ..... + n)
= 4n(n + 1)/2 using the arithmetic series formula
= 2n(n + 1)
So prove by induction: 4 + 8 + 12 + 16 + ..... + 4n = 2n(n + 1)
For n = 1
LHS = 4
RHS = 2 x 2 = 4
LHS = RHS, hence statement true for n = 1
Assume it is true for n = k
4 + 8 + 12 + 16 + ..... + 4k = 2k(k + 1)
Need to prove it is true for n = k + 1
4 + 8 + 12 + 16 + ..... + 4k + 4(k + 1) = 2(k + 1)(k + 2)
LHS = 4 + 8 + 12 + 16 + ..... + 4k + 4(k + 1)
= 2k(k + 1) + 4(k + 1) by assumption
= 2(k + 1)(k + 2)
= RHS
If the statement is true for n = k, it is also true for n = k + 1
Since it is true for n = 1, it is true for all positive integers n by induction


Q10 is also a lame one...
Prove: a + (a + d) + (a + 2d) + ....... + (a + (n - 1)d) = n(2a + (n - 1)d)/2
For n = 1
LHS = a
RHS = 2a/2 = a
LHS = RHS, hence statement true for n = 1
Assume it is true for n = k
a + (a + d) + (a + 2d) + ....... + (a + (k - 1)d) = k(2a + (k - 1)d)/2
Required to prove it is true for n = k + 1
a + (a + d) + (a + 2d) + ....... + (a + (k - 1)d) + (a + kd) = (k + 1)(2a + kd)/2
LHS = a + (a + d) + (a + 2d) + ....... + (a + (k - 1)d) + (a + kd)
= k(2a + (k - 1)d)/2 + (a + kd) by assumption
= (2ak + k(k - 1)d)/2 + (a + kd)
= (2ak + k(k - 1)d + 2a + 2kd)/2
= (2a(k + 1) + d[k(k - 1) + 2k)/2
= (2a(k + 1) + d[k² + k])/2
= (2a(k + 1) + kd[k + 1])/2
= (k + 1)(2a + kd)/2
= RHS
If the statement is true for n = k, it is also true for n = k + 1
Since it is true for n = 1, it is true for all positive integers n by induction


For 11: 5ⁿ - 1 is divisible by 4.
For n = 1
5 - 1 = 4 x 1
Therefore true for n = 1
Assume the truth of the statement for n = k
5^k - 1 = 4M (for some integer M)
Required to prove it is also true for n = k + 1
i.e. 5^{k + 1} - 1 = 4N (for some integer N)
LHS = 5^{k + 1} - 1
= 5.5^k - 1
= 5(4M + 1) - 1 by assumption (5^k - 1 = 4M => 4M + 1 = 5^k)
= 20M + 4
= 4(5M + 1)
= 4N (where N = 5M + 1 which is an integer)
= RHS
If the statement is true for n = k, it is also true for n = k + 1
Since it is true for n = 1, it is true for all positive integers n by induction

 

[jellybelly59]

that question is wrong

lol i just took it from the jones and couchman book but kk then.

 

[Trebla]

that question is wrong

It's actually correct. I think you've just read it the wrong way. The -1 bit is not part of the superscript.

 

[shantu1992]

It's actually correct. I think you've just read it the wrong way. The -1 bit is not part of the superscript.

that's it

 

[jellybelly59]

ty trebla

 

[chantszhin]

where there is pho and sugar cane drinks ....LOL.......cabramatta?