Math Help

Posts Tagged ‘calculus’

Question: Math sequences help? Multiple choice?

Find Sn for the given geometric series. Round answers to the nearest hundredth, if necessary

a1=.13, a5=1300, r=10
A. 141,059.01
B. 1,444.43
C. 90.65
D. 1,181.83

a1=-6, a5=-3584, r=4
A. –2,400
B. –36,030
C. –16,806
D. –12,606

Answer: The formula for the partial sum is Sn = a1(1 – r^n)/(1 – r), so:

1.) Sn = a1(1 – r^n)/(1 – r)
Sn = 0.13(1 – 10^5)/(1 – 10)
Sn = 1,444.43
B. 1,444.43 <===ANSWER

2.) Sn = a1(1 - r^n)/(1 - r)
Sn = -6(1 - 4^5)/(1 - 4)
Sn = -2,406
A. -2,406 <===ANSWER

Math Skills & Equations : Solving Math Sequences

Question: HELP! math word probs, sequences & series!!?

1. how many integers between 50 and 500 are divisible by 7?

2. find the sum of the powers of 5 between 5 and 9765625, inclusive.

3. find the sum of the positive 3-digit odd numbers.

(how do i set up #1? for #2, i used the geometric sum formula, and i found out the rate was 5, but the time i did out the formula it ‘overflowed’ in my calculator…did i do it wrong? and for #3, i set it up so that i found asub1, asub2, asub3 (101, 103, 105) and found “d” to make the rule, then i used the arithmetic sum formula…but the answer i got isn’t right).

any help would be great.

Answer: Q 1)

64

7 * 71 is closest to 500 yet less
7*7 is 49 which is less than 50 so 7 ints. don’t work
71-7=64

Q2) done by hand overflowed calculator

12203903905

5^1+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9
+5^10

Q3)( I forgot 0 but now fixed)

247500

101 to 999 added = ((900/2)*(101+999))/2

Sequences & Series: Summation (Sigma) Notation

Question: Help With Math integral story problem?

The book states:

It is estimated that T years from now the value of a certain parcel of land will be increasing at the rate of V’(T) dollars per year. Find an expression for the amount by which the value of the land will increase during the next 5 years.

If you could show me how you got to your answer, that would be great. Thanks!

Answer: Since V ‘(t) is the rate of increase, the actual amount of increase will be an integral of V ‘ (t). You can derive the answer by appealing to Riemann sums. Not sure what this particular author is after. If today corresponds to T=0, then the amount of increase during the next five years would be

integral(T=0 to 5) V ‘ (T) dT = V(5) – V(0)

where the last term on the right is due to the fundamental theorem of calculus.

To derive this result from “first principles” (if that is a good way to put it), we can partition the next five years—the interval from T=0 to T=5—into n small time increments of length δT (delta T). This partition would give us the times

T0=0,
T1=0 + δT = δT,
T2 = T1+δT = 2δT
…
Tn = n δT.

On each little segment, we could estimate that the rate is constant, say V ‘(T) = V ‘ (Ti). The amount of increase over each time interval would be the product of the rate times the amount of time

V ‘ (Ti) δT.

The total amount of increase over 5 years is the sum of all of these

Σ V ‘ (Ti) δT summing over i = 1 to n.

Take the limit as δT becomes infinitesimal, and you get the integral above.

Calculus Help: Integrals 1