Question

Prove these questions using Principle of Mathematical Induction ?

Answer 1

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Problem 1:
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Induction Proof - Hypothesis

We seek to prove that:

`1^2 + 3^2 + ... + (2n-1)^2 = (4n^3-n)/3 \ \ \ \ AA n in NN`

Or, equivalently:

`sum_(r=1)^n (2r-1)^2 = (n(4n^2-1))/3 \ \ \ \ AA n in NN` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` the given result gives:

`LHS = 1^2 = 1`
`RHS =(4-1)/3 = 1`

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`1^2 + 3^2 + ... + (2m-1)^2 = (4m^3-m)/3` ..... [B]

Consider the LHS of [A] with the addition of the next term, in which case we have

`LHS = 1^2 + 3^2 + ... + (2m-1)^2 + (2(m+1)-1)^2`
`\ \ \ \ \ \ \ \ = 1^2 + 3^2 + ... + (2m-1)^2 + (2m+2-1)^2`
`\ \ \ \ \ \ \ \ = 1^2 + 3^2 + ... + (2m-1)^2 + (2m+1)^2`
`\ \ \ \ \ \ \ \ = (4m^3-m)/3 + (2m+1)^2 \ \ \ \` (using [B])
`\ \ \ \ \ \ \ \ = 1/3((4m^3-m) + 3(4m^2+4m+1) )`
`\ \ \ \ \ \ \ \ = 1/3( 4m^3-m + 12m^2+12m+3 )`
`\ \ \ \ \ \ \ \ = 1/3( 4m^3 + 12m^2+12m-m+3+1-1 )`
`\ \ \ \ \ \ \ \ = 1/3( 4m^3 + 12m^2+12m+4-m-1 )`
`\ \ \ \ \ \ \ \ = 1/3( 4(m^3 + 3m^2+3m+1)-(m+1) )`
`\ \ \ \ \ \ \ \ = ( 4(m+1)^3-(m+1) )/3`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`1^2 + 3^2 + ... + (2n-1)^2 = (4n^3-n)/3 \ \ \ \ AA n in NN` QED

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Problem 2:
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Induction Proof - Hypothesis

We seek to prove that:

`1^2 -2^2+ 3^2 - ... + (-1)^(n+1)n^2 = (-1)^(n+1)(n(n+1))/2 \ \ \ \ AA n in NN`

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` the given result gives:

`LHS = 1^2 = 1`
`RHS =(-1)^2(1)(2)/2 = 1`

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`1^2 -2^2+ 3^2 - ... + (-1)^(m+1)m^2 = (-1)^(m+1)(m(m+1))/2` ..... [B]

Consider the LHS of [A] with the addition of the next term, in which case we have

`LHS = 1^2 -2^2+ 3^2 - ... + (-1)^(m+1)m^2 + (-1)^((m+1)+1)(m+1)^2`
`\ \ \ \ \ \ \ \ = (-1)^(m+1)(m(m+1))/2 + (-1)^(m+2)(m+1)^2` using [B]
`\ \ \ \ \ \ \ \ = (-1)^(m+1)(m+1)/2 { m - 2(m+1)}`
`\ \ \ \ \ \ \ \ = (-1)^(m+1)(m+1)/2 { (-1)(2(m+1)-m) }`
`\ \ \ \ \ \ \ \ = (-1)^(m+2)(m+1)/2 { 2m+2-m }`
`\ \ \ \ \ \ \ \ = (-1)^(m+2)(m+1)/2 { m+2 }`
`\ \ \ \ \ \ \ \ = (-1)^(m+2) ( (m+1) { (m+1)+1 } )/2`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`1^2 -2^2+ 3^2 - ... + (-1)^(n+1)n^2 = (-1)^(n+1)(n(n+1))/2 \ \ \ \ AA n in NN` QED