Hwk helpp p thanks in advance

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In summary, the conversation is about someone seeking help with their homework on mathematical induction. They ask for examples and for the steps to be shown clearly. Four examples are given, including proving a formula and solving an equation. The conversation ends with a reminder to read the homework guidelines and a request for ideas.
  • #1
evaboo
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hwk helpp please helpp! thanks in advance

"Prove each of the following using Mathematical induction;" show all steps
pleasee someone help.. i have a test on this tommorow and i just need some examples.. could you also try to show all steps including the words so i understand how you got there? thakns so much in advance~!

1. -1/2, -1/4, -1/8... -1/2^n = (1/(2^n))-1

2. a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

3. 1^3 + 2^3 + 3^3... + n^3 = (n^2(n+1)^2)/(4)

4. show that (3^(4n))-1 is dividislbe by 80 for all positive integral values of n
 
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  • #2
Well,did u read the HW section guidelines ?We don't do hw-s here,we only help people do their homeworks,if they get stuck.

But u haven't even started.How about some ideas...?

Daniel.

EDIT:Please,DO NOT DOUBLE POST ! :grumpy:
 
  • #3


Hello, let me help you with these problems using mathematical induction.

1. Base Case: For n = 1, we have -1/2 = (1/2) - 1 = -1/2. This is true.

2. Inductive Hypothesis: Assume that the formula holds for some positive integer k.

3. Inductive Step: We need to show that the formula holds for k+1.

a + (a+d)+(a+2d)+...+[a+(k-1)d]+[a+kd] = (k/2)[2a+(k-1)d]+[a+kd] (by inductive hypothesis)
= (k/2)[2a+kd] (since [a+kd] = 2a+kd)
= (k/2)[2a+(k+1)d]
= [(k+1)/2][2a+(k+1)d]

Therefore, the formula holds for k+1, and by mathematical induction, it holds for all positive integers n.

3. Base Case: For n = 1, we have 1^3 = (1^2(1+1)^2)/(4) = 1. This is true.

4. Inductive Hypothesis: Assume that the formula holds for some positive integer k.

5. Inductive Step: We need to show that the formula holds for k+1.

1^3 + 2^3 + 3^3... + k^3 + (k+1)^3 = (k^2(k+1)^2)/4 + (k+1)^3 (by inductive hypothesis)
= (k^4 + 2k^3 + k^2)/4 + (k+1)^3
= (k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4)/4 (by expanding (k+1)^3)
= (k+1)^2(k+2)^2/4

Therefore, the formula holds for k+1, and by mathematical induction, it holds for all positive integers n.

4. Base Case: For n = 1, we have (3^(4*1))-1 = 80 which is divisible by 80. This is true.

5. Inductive Hypothesis: Assume that the formula
 

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