HELP with these question (1 Viewer)

[Mellonie]

1. How many terms of the serious 9 + 18 + 36 + ... are needed to gve sum of 1143




2. n=1 Signma notation n=8 16(.75)^(n-1) {without writing the whole thing out from n=1 to n=8 like it can be done with just sum of n terms formula right..}



Their easy but i dunno y i'm not getting the answer... maybe the book is wrong so yer check it for me plz

 

[skyrockets1530]

How many terms of the serious 9 + 18 + 36 + ... areneeded to gve sum of 143

n=1 Signma notation n=8 16(.75)^(n-1) {without writing the whole thing out from n=1 to n=8 like it can be done with just sum of n terms formula right..}



Their easy but i dunno y i'm not getting the answer... maybe the book is wrong so yer check it for me plz

S = [a(rⁿ - 1)]/(r - 1)
r = 2, a=9, S=143
143 = [9(2ⁿ - 1)]/(2 - 1)
143 = 9(2ⁿ - 1)
143/9 = 2ⁿ - 1
152/9 = 2ⁿ
ln(152/9) = nln2
n=4.08
you cant really have a partial term of a series, but that seems to be right

 

[Mellonie]

yer hey thanx i figured it out.. my formula was wrong

 

[word.]

because 143/9 + 1 = 152/9...

q2.
n=8
Sigma 16(.75)^(n-1)
n=1

= 16 + 16(0.75) + ... + 16(0.75)⁷

a = 16, r = 0.75, n = 8
etc

 

[icycloud]

1. How many terms of the serious 9 + 18 + 36 + ... are needed to gve sum of 1143

This is a geometric series with first term of 9 and common ratio of 2. Thus,

a = 9, r = 2

S_n = a(rⁿ-1) / (r-1)
= 9(2ⁿ-1)
>= 143

Thus, 9(2ⁿ-1) >= 143
2ⁿ >= 143/9 + 1
n ln(2) >= ln(152/9)
n >= ln(152/9) / ln(2)
>= 4.07800.......

But n is an integer, therefore, n = 5
Thus, 5 terms are necessary for the sum to approach 143.

2. n=1 Signma notation n=8 16(.75)^(n-1) {without writing the whole thing out from n=1 to n=8 like it can be done with just sum of n terms formula right..}

See word.'s solution above.