sequences/binomial 問題！ (20)

6.Find the sum of the infinite GS (2/3)– (4/9) + (8/27) – (16/81) + …… (打5到份數）

Ans:

T(1) =2/3
T(2)= -4/9 = (2/3)(-2/3)
T(3)=8/27 =(2/3)(-2/3)^2
...
T(n) =(2/3)(-2/3)^(n-1)

so the sum of GP = a/ (1-R) =(2/3)[1-(-2/3)] =2/5 //

7.Let S n = 2 + 6 + 18 + …… + 2 x 3^(n-1)

Find the Least value of n for which S n > 150,000

S(n) >15000
=> a(R^n-1)/(R-1) >15000
=> 2[3^n-1]/[3-1] >15000
=>3^n-1>15000
=>log 3^n > log 15001
=> n > log 15001 / log 3 = 8.8

so least value of n is 9 .

5.5. For the GS 2, -6, 18, …… find the seventh term ad the sum of the first ten terms.

Ans:
T(1) =2 , T(2) = -6 =(2)(-3) ,T(3) =18 =(2)(-3)^2 ....
T(n) = (2)(-3)^(n-1)

S(10) = a(R^n-1)/(R-1)=(2)[(-3)^10-1]/(-3-1)=-29524//

4a)
let a(n) be the distance she ran on the nth day

a(1) = 1km
a(2) = 1.25km
a(3) = 1.5km

a(n) = 0.75 + 0.25 x n = 10km
hence n = 37

on day 27, she ran 10km

4b)
a(1) + a(2) + a(3) + ... ... + a(37)
= 37 x 0.75 + 0.25 x (1 + 2 + 3 + ... ... + 37) = 203.5km

Q4a: let a(n) be the distance she ran on the nth day

a(1) = 1km
a(2) = 1.25km
a(3) = 1.5km

a(n) = 0.75 + 0.25 x n = 10km
hence n = 37

on day 27, she ran 10km

Q4b: a(1) + a(2) + a(3) + ... ... + a(37)
= 37 x 0.75 + 0.25 x (1 + 2 + 3 + ... ... + 37) = 203.5km

Q5: the common ratio R = -3

the seven term = 2 x [(-3) ^ 6] = 1458

the sum to the tenth term = 2 x {1 - [(-3) ^ 10]} / [1 - (-3)] = -29524