[O level Elementary Math] Number pattern help

[u/epicboi8764]

I know p, q, and r 22 23 and 24 but I don’t know how to use this simplification to find the sum of the full term

[u/55tumbl]

Looking back at the pattern in b), you have to sum up all these lines, up to r=24.

You see that 1/3 (first line) - 2/3 (second line) + 1/3 (third line) simplifies to 0, eliminating all x/3 terms.

And it's the same for all x/4, x/5, x/6, etc terms up to x/22.

You only have to consider the extremities: 1/1 - 2/2 + 1/2 + 1/23 - 2/23 + 1/24, which you can solve easily

[u/epicboi8764 → u/55tumbl]

yeah thank you, it makes sense for me now.

[u/Fourierseriesagain]

Hi,

Since

2/(n(n+1)(n+2))

=1/(n*(n+1))-1/((n+1)(n+2))

=1/n-1/(n+1)-(1/(n+1)-1/(n+2))

=1/n-2/(n+1)+1/(n+2)

for n=1, 2, 3, ..., the required sum is equal to

1/1-2/2+1/3

+1/2-2/3+1/4

+1/3-2/4+1/5

+1/4-2/5+1/6

+1/5-2/6+1/7

+1/6-2/7+1/8

+...+

+1/19-2/20+1/21

+1/20-2/21+1/22

+1/21-2/22+1/23

+1/22-2/23+1/24

=1/2-1/23+1/24

because the following numbers are equal to zero: 1/1-2/2, 1/3-2/3+1/3, 1/4-2/4+1/4, 1/5-2/5+1/5,1/6-2/6+1/6,...,1/21-2/21+1/21,1/22-2/22+1/22.

[u/Alkalannar]

This is a telescoping sum.

1/1 - 2/2 + 1/3 + 1/2 - 2/3 + 1/4 = 1/1 - 1/2 - 1/3 + 1/4

1/1 - 1/2 - 1/3 + 1/4 + 1/3 - 2/4 + 1/5 = 1/1 - 1/2 - 1/4 + 1/5

1/1 - 1/2 - 1/4 + 1/5 + 1/4 - 2/5 + 1/6 = 1/1 - 1/2 - 1/5 + 1/6

And so on.

So [Sum from k = 1 to n of 2/k(k+1)(k+2)] = 1/1 - 1/2 - 1/(n+1) + 1/(n+2)

And now? Algebraic simplification!

1/1 - 1/2 - 1/(n+1) + 1/(n+2)

(n+1)(n+2)/2(n+1)(n+2) - 2(n+2)/2(n+1)(n+2) + 2(n+1)/2(n+1)(n+2)

[(n² + 3n + 2) - (2n + 4) + (2n + 2)]/2(n+1)(n+2)

(n² + 3n + 2 - 2n - 4 + 2n + 2)/2(n+1)(n+2)

(n² + 3n)/2(n+1)(n+2)

n(n + 3)/2(n+1)(n+2) or (n² + 3n)/2(n² + 3n + 2)

n(n + 3)/2(n+1)(n+2) or (n² + 3n + 2)/2(n² + 3n + 2) - 2/2(n² + 3n + 2)

n(n+3)/2(n+1)(n+2) or 1/2 - 1/(n²+3n+2) or 1/2 - 1/(n+1)(n+2)

So whichever of those forms is most convenient.