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junior college 1 | H2 Maths
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Stuck for part iii
m + 1 = 3
m = 2
Setting r = 1 and letting mr + 1 = 4,
m + 1 = 4
m = 3
There are basically two series in there. You'll need to regroup the terms first.
1/1·3 + 1/1·4 + 1/3·5 + 1/4·7 + 1/5·7 + 1/7·10 + ... 1/39·41 + 1/58·61
= 1/1·3 + 1/3·5 + 1/5·7 + ... 1/39·41
+ 1/1·4 + 1/4·7 + 1/7·10 + ... 1/58·61
= 1/(2·1 + 1 - 2)(2·1 +1) + 1/(2·2 + 1 - 2)(2·2 + 1) + 1/(2·3 + 1 - 2)(2·3 + 1) + ... 1/(2·20 + 1 - 2)(2·20 + 1)
+ 1/(3·1 + 1 - 3)(3·1 + 1) + 1/(3·2 + 1 - 3)(3·2 + 1) + 1/(3·3 + 1 - 3)(3·3 + 1) + ... 1/(3·20 + 1 - 3)(3·20 + 1)
□20□□□□ □□ □□□□□□20
= ∑ 1/(2r + 1 - 2)(2r + 1) + ∑ 1/(3r + 1 - 3)(3r + 1)
r=1□□□□□□□□□□□□ r=1
= 20/(2·20 + 1) + 20/(3·20 + 1)
= 20/41 + 20/61
= 2040/2501
The trick is to spot that there is a common difference of 2 for the products in the denominator of the odd terms and common difference of 3 for that of the even terms, from which you can infer that there are two series with different values of m.
Also, 41 = 2·20 + 1 and 60 = 3·20 + 1 follows the mr + 1 and tells you that each series ends at 20 terms (i.e n = 20)
Unfortunately, on my phone, it appears that the second 20 is placed some distance to the right of where it should be. I guess it’s just different phones and different computer screens.
The best alternative was these white square boxes.
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