Sequences using Sigma

[mattgad]

Hi all, I have this question

For what value of n does:

n

Sigma (5r + 3)

r = 1

first exceed 1000?

I've scanned in my book, if its any easier to read the question.

I have tried to tackle this question about 4 times now, but still cant get my head around it. Any advice on how to tackle it will be greatly appreciated. Thanks.

 

[pka]

The sum is equal to 5∑r + ∑3=5[n(n+1)/2]+3n.
Find the least n that makes 5[n(n+1)/2]+3n >1000

 

[tkhunny]

n = 1 [5(1) + 3]
n = 2 [(5*1)+3)] + [(5*2) + 3] = 5*3 + 3*2
n = 3 [(5*1)+3)] + [(5*2) + 3] + [(5*3) + 3]= 5*6 + 3*3

With each additional term, '3' is added and some number of fives. Let's examine the '5's.

1 + 2 + 3 + ... + n This is a common sequence of terms. It's sumis well known (n*(n+1))/2. Try it.

(1*2)/2 = 1 -- [5(1) + 3*1] Yup.
(2*3)/2 = 3 -- [5(3) + 3*2] Yup
(3*4)/2 = 6 -- [5(6) + 3*3] Yup

n terms, then, totals, 5*[n*(n+1)/2] + 3*n

Find out what it takes to make that expression greater than 1000.

 

[mattgad]

Thank you both for the help, that equation makes sense when I subsitute my numbers in, but how did you get to it? How did you get to the (n*(n+1))/2 and times by 5? Is that the value of r?

 

[stapel]

These are formulas that should have been covered in the text and/or in class.

. . . . .sum_{[i=1 to n]} i = [n(n + 1)]/2

. . . . .sum_{[i=1 to n]} a = an

...where "a" is any constant value.

Check your text and notes; you are likely expected to have these sums memorized, so you can use them in other exercises (such as this), so you'll want to know them before the test.

Eliz.

 

[soroban]

Hello, mattgad!

Write out the terms: .8 + 13 + 18 + 23 + ... + (5n+3)

We have an Arithmetic Series with first term a = 8, common difference d = 5, and n terms.

. . The sum is: .S_n .= .(n/2)[2a + d(n-1)]

So we have: .Sn .= .(n/2)[2·8 + 5(n-1)] .> .1000

. . or the quadratic inequality: .5n² + 11n - 2000 .> .0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._____
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . -11 ± √40121
Solving the <u>equation</u>, we get: .n .= . ----------------- . = . 18.93, -105.65
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Since the inequality holds for n > 18.93, then: .n = 19.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

n = 18: .S₁₈ .= .(18/2)[2·8 + 5(17)] .= .909
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . YAY! . It's Miller time!
n = 19: .S₁₉ .= .(19/2)[2·8 + 5(18)] .= .1007
.