Adapting Alison's Method
∑ 100 i=1 i
= 1+2+3+⋯+98+99+100
100+99+98+⋯+3+2+1
All pairs equal 101. There are 100 terms so 101x100=10100
10100/2=5050
Therefore ∑ 100 i=1 i = 5050
2+4+6+⋯+96+98+100
can also be written as ∑ 50 i=1 2i
= 2+4+6+⋯+96+98+100
100+98+96+⋯+6+4+2
All pairs equal 102. There are 50 terms so 102x50=5100
5100/2=2550
Therefore ∑ 50 i=1 2i = 2550
∑ 20 k=1 (4k+12)
= 16+20+24+⋯+84+88+92
92+88+84+⋯+24+20+16
All pairs equal 108. There are 20 terms so 108x20=2160
2160/2=1080
Therefore ∑ 20 k=1 (4k+12) = 1080
37+42+47+52+⋯+102+107+112
can also be written as ∑ 22 k=7 (5k+2) meaning it has 16 terms. (22-6=16)
∑ 22 k=7 (5k+2)
= 37+42+47+⋯+102+107+112
112+107+102+⋯+47+42+37
All pairs equal 149. 149x16=2384
2384/2=1192
Therefore 37+42+47+52+⋯+102+107+112=1192
The sum of the first n terms of the sequence a,(a+d),(a+2d),(a+3d)…
This was the proof of the formula for an arithmetic progression.
∑ n d=0 (a+d)
= a + (a+d) + (a+2d)+⋯+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d)
(a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+⋯+(a+2d) + (a+d) + a
All pairs equal 2a+(n-1)d
[n terms x (2a+(n-1)d)]/2 = (n/2)(2a+(n-1)d) which is the rule for a sum of an arithmetic progression.
Using this rule for 17+21+25+…>1000
a (the first term)=17, and d (the constant added)=4
Therefore (n/2)[(2x17)+4(n-1))]>1000
simplifying this equation:
(n/2)(30+4n)>1000
which is a quadratic: 2n^2+5n+1000>0
using the quadratic equation, n=18.92 (2 d.p) which would be rounded up because the summation must be greater than 1000, therefore n=19.
If the number is not divisible by 2 or 3 then it can't be divisible by any number with 2 or 3 as a factor. Therefore only prime numbers (excluding 2 and 3) and multiples of 5 which are not divisible by 3.
1 + 5 + 7 + 11+⋯+899+991+995+997
997+995+991+899+⋯+11+ 7 + 5 + 1
which alternate between equalling 998 and 1000
To work out how many terms there were in the summation, I worked out the rule for the nth term, using trial and error:
If n is even the rule is 3n-1
If n is odd the rule is 3n-2
3n-2=997 therefore n=333, meaning 997 is the 333rd term and there are 333 multiples of 998.
33n-1=995 therefore n=332, meaning 995 is the 332nd term and there are 332 multiples of 1000.
(333x998)+(332x1000)=664334
664334/2=332167
Using the rule from the summations previously:
32=[n(f+l)]/2, where n is the number of terms, f is the first term and l is the last term.
64=n(f+l)
so n and (f+l) must be factors of 64.
Factors of 64: 1,2,4,8,16,32,64
Using a proof by exhaustion:
2 consecutive integers (n terms)=32 (f+l), cannot be done because odd+even=odd Therefore not a factor of 64 because 64 has no odd factors (excluding 1)
4 consecutive integers (n terms)=16 (f+l)
4 consecutive integers therefore l-f=3. There are no positive integers with a difference of 3 that sum to 16.
8 consecutive integers (n terms)=8 (f+l)
8 consecutive integers therefore l-f=7. There are no positive integers with a difference of 7 that sum to 8.
This is because a sum of an even number (in this case (f+l): a factor of 64) must be created by odd+odd or even+even. Therefore f and l must be both odd or both even (meaning the difference between them would be even). Therefore there would be an odd number of integers, which would mean n would be odd and as 64 has no odd integers, n cannot be odd.
Therefore no set of consecutive positive integers whose sum is 32.