Please help me find out what group of sequence the number 2678 is in.

F

Felix_GG

New member
Joined
May 4, 2023
Messages
20
1683299222391.png

I'm lost, i was sure that i was super close to find the correct answer, even all 4s in Inequality did erase. After that i tried to calcute what's power is 1339, and it was between 36 and 37 and i thought 2678 is in 37th group but that's not the right answer. The right answer for
52
and for
77
. 1683299726869.png

Where is my mistake? Sorry for my poor english.
 
Dr.Peterson said:
Please tell us your answer for part 1, and explain briefly in words what you are doing here. Where did this inequality come from?
Click to expand...
The answer for the first part is 0, 1, -2 and 2, so n^2-2n+2.

I was solving the second part in the way YouTube tutorials showed me.

The first step is to let 2678 be in n-th group, after that I took the equation for sum of (n-1)-th group and figured, that 2678 should be larger than the sum but be less than the sum of n-th group, that's what I wrote.
Equation with sigma symbol is the sum of (n-1)-th group.
 
Felix_GG said:
The answer for the first part is 0, 1, -2 and 2, so n^2-2n+2.

I was solving the second part in the way YouTube tutorials showed me.

The first step is to let 2678 be in n-th group, after that I took the equation for sum of (n-1)-th group and figured, that 2678 should be larger than the sum but be less than the sum of n-th group, that's what I wrote.
Equation with sigma symbol is the sum of (n-1)-th group.
Click to expand...
Did you not consider using the answer to the first part as the basis for your work on the second?

The nth group starts with n^2-2n+2; the next group starts with (n+1)^2-2(n+1)+2. You want to find n such that 2678 is between those two. You didn't need to start over with a new summation formula (which seems to be wrong).

At least, rethink what you want to sum, or whether you need to do so at all. (Did you solve the first part as a similar sum? Your answer is right, but you didn't show us how you got it.) And you should check whether your summation formula works for the first few terms before doing the hard part.
 
Dr.Peterson said:
Did you not consider using the answer to the first part as the basis for your work on the second?

The nth group starts with n^2-2n+2; the next group starts with (n+1)^2-2(n+1)+2. You want to find n such that 2678 is between those two. You didn't need to start over with a new summation formula (which seems to be wrong).

At least, rethink what you want to sum, or whether you need to do so at all. (Did you solve the first part as a similar sum? Your answer is right, but you didn't show us how you got it.) And you should check whether your summation formula works for the first few terms before doing the hard part.
Click to expand...
Wow, I finally understood my mistake. I wrongly assumed that 2678 was the number of term, when it's actually the value of the term. Eventually it was just the language issue. I got all answers. You were right, i just needed to use the formula from the first part, i don't know why i have tried to figure a new formula, when i already had the correct one. From now on i will be more cautious. You have led me to the right answer, thanks a lot! 1683470752295.png
 
Felix_GG said:
Wow, I finally understood my mistake. I wrongly assumed that 2678 was the number of term, when it's actually the value of the term. Eventually it was just the language issue. I got all answers. You were right, i just needed to use the formula from the first part, i don't know why i have tried to figure a new formula, when i already had the correct one. From now on i will be more cautious. You have led me to the right answer, thanks a lot! View attachment 35744
Click to expand...
It looks like you solved this by trial and error using the formula; you can also use the quadratic formula, solving for n to make the formula equal 2678, and rounding down (then checking).
 
You must log in or register to reply here.
Top