Fairly simple induction qu i can't work out (1 Viewer)

[VenomP]

Prove by mathematical induction that

12/(1x3x4) + 18/(2x4x5) + 24/(3x5x6) + ... + [6(n+1)]/[n(n+2)(n+3)] = 17/6 - 1/(n+1) - 1/(n+2) - 4/(n+3)

I'm seriously so bad at these.

 

[tommykins]

it'd probably be easier if you made the RHS into a common denominator

 

[Drongoski]

Prove by mathematical induction that

12/(1x3x4) + 18/(2x4x5) + 24/(3x5x6) + ... + [6(n+1)]/[n(n+2)(n+3)] = 17/6 - 1/(n+1) - 1/(n+2) - 4/(n+3)

I'm seriously so bad at these.

Proof plan goes like this:

1) Show true for n = 1
2) Assume true for n = k (for k >= 1)
3) Then show true for case of n = k+1

Proof:

For n = 1 LHS = 1st term = 12/(1x3x4) = 1
RHS = 17/6 - 1/2 - 1/3 - 4/4 = 1
Thus true for n = 1

Now assume true for n = k >= 1
i.e. 12/(1x3x4) + 18/(2x4x5) + . . . + 6(k+1)/(k(k+2)(k+3)) = 17/6 - 1/(k+1) - 1/(k+2) - 4/(k+3)

[Now we need to show that it is true for the next value of n, i.e. for n = k+1
But what does this mean?? It means the "formula" is true when 'k' is replaced by 'k+1'; this means where we had k+1 before, it is now k+2, where we had k+2 before it is now k+3 etc ]

Now for n = k+1, LHS =

12/(1x3x4) + 18/(2x4x5) + . . . + 6(k+1)/(k(k+2)(k+3)) + 6(k+1 + 1))/((k+1)(k+1 + 2)(k+1 + 3))
= 17/6 - 1/(k+1) - 1/(k+2) - 4/(k+3) + 6(k+2)/((k+1)(k+3)(k+4))
= 17/6 - 1/(k+2) - 1/(k+3) + { 6(k+2)/((k+1)(k+3)(k+4)) - 1/(k+1) - 3/(k+3) }
= 17/6 - 1/(k+2) - 1/(k+3) - { 4(k^2 + 4k +3)/(k+1)(k+3)(k+4)) }
= 17/6 - 1/(k+2) - 1/(k+3) - 4/(k+4) [ since k^2 + 4k + 3 = (k+1)(k+3) ]
= The RHS with n = k+1

We have now shown formula is true for n = k+1

Thus, by the Principle of Mathematical Induction, formula is true for all n = 1,2,3,4 . . .

[Apologies to viewers who are super bright. I've shown the solution in some painful detail + comments along the way to help those who are struggling with Math. Induction problems. Hopefully my attempt here has been of some help.]

 

[Drongoski]

VenomP - Do you understand now ??

 

[lychnobity]

Ahh, you could have used latex. It's painful to read regardless of brightness

 

[Drongoski]

Ahh, you could have used latex. It's painful to read regardless of brightness

I know. I don't know how to use LaTeX. Maybe you can teach me. Or I should just give up making any further contributions. I find it painful not being able to use LaTex. It's so embarrassing.

 

[tommykins]

I know. I don't know how to use LaTeX. Maybe you can teach me. Or I should just give up making any further contributions. I find it painful not being able to use LaTex. Makes me look dumb.

so does your small size font -_-