Question

sequences

You need to first recognize that the sequence is:

1/1

1/4
3/4

1/9
3/9
5/9

1/16
3/16
5/16
7/16

For every m, the terms 1, 3, 5...,2m-1 are numerators, all divided by m^2.

Denominator 1: a(1)
Denominator 4: a(2), a(3)
Denominator 9: a(4), a(5), a(6)
Denominator 16: a(7), a(8), a(9), a(10)
Denominator 25: a(11), a(12), a(13), a(14), a(15)
Denominator 36: a(16) to a(21)
Denominator 49: a(22) to a(28)
Denominator 64: a(29) to a(36)
Denominator 81: a(37) to a(45)
Denominator 100: a(46) to a(55)
etc..

You might notice in each case the range ends on a triangular number.

So for part (a), it is simply:

a(46) = 1/100
a(47) = 3/100
a(48) = 5/100
a(49) = 7/100
a(50) = 9/100

For part (b), you probably ought to group them up:

1/1 + (1+3)/4 + (1+3+5)/9 + (1+3+5+7)/16 + ... + (1+3+5+7+9)/100

Note that we stop at a(50), otherwise that term with denominator 100 would be larger.

In each case except the final term, you get 1. Why? Because there is an identity:

1 + 3 + 5 + ... + (2m+1) = m^2

That means your total summation is nine 1s and what's there over 100:

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1+3+5+7+9)/100 = 9 + 25/100 (or 9.25 or 925/100)

For part (c), the largest in any denominator group is the term (2m+1)/m^2.

You want the largest value of m so that:

(2m+1)/m^2 ≥ 1/10

You can just rearrange that to get:

20m + 10 ≥ m^2

And you can either try a bunch of values of m until it doesn't work, or use the quadratic formula to find the roots of:

20m + 10 = m^2

For that you get 10+√(110), which is slightly more than 20, so the answer is the term where m=20, so (2*20+1)/20^2, which is 41/400. That is just slightly more than 1/10.

When m=21, you'll see that (2*21+1)/21^2 is 43/440, slightly less than 1/10.

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#### Clyde Oliver

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