Find the sum of each series by using appropriate formulas. Show your work.
MY WORK:
(N is the last number in the sequence)
1+2+3+...+230
n(n+1)/(1*2)
4+8+12+...+360
n(n+4)/)4*2)
1+3+5+...+123
((n+1)/2))^2
The problem:
53+59+65+...+233
n(n+6)/(6*2)
233(239)/12 = decimal???
There can't be a decimal when finding a sum of a sequence. What did I do wrong?
d = 6
233 - 53 = 180 ... so, 31 terms
s = (233 + 53) * 31 / 2
I have no idea what that first part of your post is supposed to be.
your problem:
53+59+65+...+233
We have to know how many terms.
a=53, d=6, n= ?
term(n) = a + (n-1)d
233 = 53 + 6(n-1)
180 = 6n - 6
6n = 186
n = 31
sum(31) = (31/2)(first + last)
= (31/2)(53+233) = 4433
which is what Scott's answer would give you
To find the sum of the series 53+59+65+...+233, you correctly identified that the formula to use is n(n+6)/(6*2). However, it seems like you made a calculation error when plugging in the values.
Let's go through the calculation step by step to identify the mistake:
- Identify the first and last terms of the series:
First term (a): 53
Last term (b): 233
- Calculate the number of terms in the series:
Number of terms (n) = (b - a)/6 + 1
= (233 - 53)/6 + 1
= 180/6 + 1
= 30 + 1
= 31
- Now, calculate the sum by using the formula:
Sum = n(a + b)/2
= 31(53 + 233)/2
= 31(286)/2
= 8834/2
= 4417
Therefore, the sum of the series 53+59+65+...+233 is 4417. You made a mistake in calculating the number of terms, which led to an incorrect result.