Find the sum of -8/5+-2/3
Answers
The answer to your question is -34/15.
Answer:
If n is preceded by “-”:
1–2+3–4+5–6+7-…-n
(1–2)+(3–4)+(5–6)+(7–8)…((n-1)-n) = ?
If n = 1, the answer is 1.
If n = 2, then we have 1–2 = -1.
If n = 3, then we have -1+3 = 2
if n = 4, then we have 2–4 = -2
if n = 5, then we have -1 + -1 + 5 = -2 + 5 = 3
if n = 6, then we have 3 - 6 = -3
Let’s see if we can find a pattern in the above (other answers have already revealed the tendency of the summation depending on -n or +n, but I don’t think that that was your question).
Since I’ve specified that we’re currently looking at the case where the sequence ends in -n, let’s only look at those situations.
We know that (by looking at the sequence), the sequence will end in -n is n is an even number (look up there: -2…-4…-6…-n)
So, saying that this sequence ends in -n is that same thing as saying that n is even.
Now, let’s look at those first few even examples
n = 2, sum is -1
n = 4, sum is -2
n = 6, sum is -3
So, with each successive even number, the sum decreases by 1.
Also, notice that -1 = -(2/2), -2=-(4/2), -3=-(6/2)
In other words, the sum is equal to the negative of half of n
We could write this as: if n is even, sum is -(n/2).
Now, let’s check out our solution for sequences ending in +n.
1–2+3–4+5–6+7-…+n
(1–2)+(3–4)+(5–6)+(7–8)+…+n = ?
If n = 1, the answer is 1.
If n = 2, then we have 1–2 = -1.
If n = 3, then we have -1+3 = 2
if n = 4, then we have 2–4 = -2
if n = 5, then we have -1 + -1 + 5 = -2 + 5 = 3
if n = 6, then we have 3 - 6 = -3
Let’s see if we can find a patter with the sequences ending in +n, which are those cases where n is not even (aka n is odd).
n=1, sum is 1
n=3, sum is 2
n=5, sum is 3
Notice that with each successive odd n, the sum is increased by 1.
Also notice, that the sum is equal to n minus the previous sum.
n=5, sum is 3, which is 5–2, and 2 is the sum where n=3.
n=3, sum is 2, which is 3–1, and 1 is the sum where n = 1.
Let’s test this for n = 7
We know that for n=6, sum is -3. So let’s just add 7 to -3.
7+-3 = 4.
n=7, sum is 4, which is 7–3, and 3 is the sum where n = 5.
Alright, so how do we simplify this and make it more generally applicable?
We could write this as: if n is odd, sum = n-(sum where n = n-2)
Right?
sum = n-(sum where n = n-2)
Try 7: sum=7-(sum where n=5), sum=7-(3), sum=4.
Woohoo!
If you thought this one was neat (or maybe you thought I took shortcuts or cheated by using the sum function as part of my solution), check out recursion. It’s awesome.
NOTE: that whole sum=n-(sum where n=n-2) thing? You might be wondering where I got that n=n-2, right? That’s just short hand for “the next lowest odd number” for our purposes. So, if n=7, then n-2=5, and 5 is the next lowest odd value for n.
So, to find any specific sum for any value of n for that particular sequence you’ve specified…