Answer:
a: start at n=1 end at infinity equation = 4^n
b: start at n=3 end at 5 equation = n/(n+1)
c: start at n=1 end at 100 equation = (-1)^n*(1/n)
d: start at n=1 end at n=9 equation = -2
e: starts at n = 0, ends at infinity equation = 5 + 2n
f: starts at n = 0 ends at 4 equation = (7+n)*m^(6+n)
Let me know if you don't see how this works.
Step-by-step explanation:
So sigma notation has three parts, the start, the end and the equation, the start is below the sigma, the end is above and the equation of course is to the side. i will tell you the parts, but let me know if you need further explanation
4 + 16 + 64 + 256
I always start with checking if it is arithmetic which would mean something is added Well, you add 12 to 4 to get 16 and way more than 12 to get to 64, so that's not right.
Next I check geometric. 4*4 = 16 then 16*64, so we're good there. Geometric sequences have the form ab^x where a is the starting number and b is the number that is multiplied by. well both are 4 so we can just use 4^x, though with sigma notation n s usually used instead of x.
so, the equation is 4^n, the starting point is n=1 becaus the starting number is 4 and for 4^n to be 4 n has to be 1. Now, where does it end? Well it doesn't have one, which is shown by the ... at the end. so you should put infinity.
b. 3/4 + 4/5 + 5/6
Not arithmetic or geometric, you can't add anything to each or multiply anything 3/4 gets .05 added to it but 4/5 doesn't and 3/4 is mltiplied by 16/15 but 4/5 is not. So now we look for more special sequences..
Well for 3/4 4 is one more than 3, and the same in 4/5 and 5/6. Also, 4 is the denominator in the first then the numerator in the second, so we have a pattern. the equation is n/(n+1) then it starts at 3 and ends at 5 since there is no ...
c. -1 + 1/2 - 1/3 + 1/4 - ... + 1/100
Again, not arithmetic of geometric. First thing to notice is that it alternates from - to +, now how can a number do that? well what is a negative number to different exponents? let's take (-1)^1, (-1)^2, (-1)^3 and so on. ou'll notice this alternates between -1 and 1, so let's use that. also note the first number is negative, so we want to make sure that's the case for us.
Now for the numbers themselves. I think the pattern is pretty obvious 1/1, 1/2, 1/3 and so on, so the equation is 1/n up to 100. So this one does have an end, even though there's that ..., there is also a number after that though. So let's set this up with a start at 1 and end at 100
(-1)^n*(1/n) and it keeps the negative as the first number too
d. -2 - 2 - 2 - 2 - 2 - 2 - 2 - 2
This one's pretty easy, there are eight 2s. since the 2s have nothng to do with any variable the equation is just 2, and we can start at any n as long as we end at an n 8 away. let's do it normally and start at n=1 so we end at n=9
e. 5 + 10 + 17 + 26 + ...
... means no end again. No obvious pattern so let's check arithmetic and geometric. Well we add 5 to 5, 7 to 10 then 9. Now, those aren't the same numbers, but they are increasing 5 has 2 added to it to get to 7, then 7 gets 2 added to get to 9. so the equation is 5 + (2n) as long as it starts at n=0. So all together that's starts at n = 0, ends at infinity equation = 5 + 2n
f. 49m^6 + 64m^7 + 81m^8 + 100m^9 + 121m^10
Little bit tricky, hope you know your square numbers. more obviously though the exponents are increasing y 1, so we have that. m^(6+n) where n starts at 0 and then ends at 4 to get to m^(6+4) at the end. The numbers of the terms meanwhile are increasing as well. Hopefully you can tell they are square numbers. 49 = 7^2 64 = 8^2 and so on. so it starts at 7^2so now we know the equation is (7+n)m^(6+n). You could also play with the start and ending points to make it look simpler, but I would just go with that.