biina:

Just an edit (hope its right)

it doesnt change anything.

iwefactria:

it doesnt change anything.

I know, I was only try to make the question clearer by removing things I felt could cause confusion. (like the periods that could be interpreted as decimal points)

at the risk of missing it i'd say from a qualitative point of view that the answer is 5050

netotse:

at the risk of missing it i'd say from a qualitative point of view that the answer is 5050

You're right.

1+2+. . .+100 = 100*101/2 = 50*101=5050

Find the sum of the following:

1+1.5+2+2.5+3+3.5+. . .+99+99.5+100

RichyBlacK:

Find the sum of the following:

1+1.5+2+2.5+3+3.5+. . .+99+99.5+100

need to include options

RichyBlacK:

You're right.

1+2+. . .+100 = 100*101/2 = 50*101=5050

you are right

RichyBlacK:

Find the sum of the following:

1+1.5+2+2.5+3+3.5+. . .+99+99.5+100

the ans= 10049.5

2(1) + (198)(0.5) = 101*199/2 = 20099/2 = 10049.5

biolabee:

2(1) + (198)(0.5) = 101*199/2 = 20099/2 = 10049.5

RichyBlacK:

You're right.
1+2+. . .+100 = 100*101/2 = 50*101=5050

how're u guys getting the formula is it using A.P (arithmetic progression)?`

netotse:

how're u guys getting the formula is it using A.P (arithmetic progression)?`

yes A.P

(Hard)
Find (as shown below) the sum of the infinite series, with the nth term defined as n*r^(n-1).

1 + 2r + 3r^2 + 4r^3 + 5r^4 + 6r^5 + 7r^6 +  . . .   . . . . . . . . . . = ?

where |r|<1

1. Infinity
2. 0
3. 1/(1-r)
4. r/(r-1)
5. 1/(1-r)^2
6. (r^2)/(1-r)^2
7. 1/(1-r^2)
8. ln(1-r)

Notes:
^ is the exponentiation function, and |x| is the absolute value of x.

I will give a hint if no one solves it in the next 24hrs.

3

I think the ans should (r^2)/(1-r)^2

biolabee:

3

iwefactria:

I think the ans should  (r^2)/(1-r)^2

Please no guessing.
If you know the answer please state so, giving a brief explanation of why. Answers with no reasoning given will be taken as a guess, and subsequently ignored (even if right).

Thanks for resurrecting my fav thread

biina:

Please no guessing.
If you know the answer please state so, giving a brief explanation of why. Answers with no reasoning given will be taken as a guess, and subsequently ignored (even if right).

rotflmao

I think the answer is infinity

this is wrong. . .(abeg no more maths now. . .i've had 6 years of it!)

*Toyinrayo:

Thanks for resurrecting my fav thread 

You are welcome.

rotflmao 
I think the answer is infinity 

You just couldn't resist the temptation  <gives up>

netotse:

this is wrong. . .(abeg no more maths now. . .i've had 6 years of it!)

sounds like you were traumatized by the experience. I can definitely relate jare, some classes just give you bad memories and cause you nightmares.

I am going to try and mix it up a bit, in terms of difficulty and scope (but its so hard to come up with questions). Will try to avoid esoteric questions in the future. The hope is that more people get involved hereby enriching the experience.

(Hard)
Find (as shown below) the sum of the infinite series, with the nth term defined as n*r^(n-1).

1 + 2r + 3r^2 + 4r^3 + 5r^4 + 6r^5 + 7r^6 +  . . .   . . . . . . . . . . = ?

where |r|<1

1. Infinity
2. 0
3. 1/(1-r)
4. r/(r-1)
5. 1/(1-r)^2
6. (r^2)/(1-r)^2
7. 1/(1-r^2)
8. ln(1-r)

Notes:
^ is the exponentiation function, and |x| is the absolute value of x.

Hint:
1. The nth term in the series above, n*r^(n-1), can be obtained by differentiating the nth term of the geometric progression (GP) series, r^n i.e. d(r^n)/dr = n*r^(n-1)
2. Differentiation is a linear operation i.e. dx+dy=d(x+y)
3. The sum of a GP series with first term a=1, and ratio r is easily found.

**Please no guessing. Answers should be followed by a brief comment explaining the reason for the choice**

i am not sure o pls correct me .
when delta G is positive.

mrCracles u are quite intelligent if i must say.THUMBS UP DUDE.

who can answer this question
a woman had 15 naira she spent 9 naira , gained 12 naira and loss 48 naira how much is left?

@2angels
biina's question hasnt been answered. . . dont u think its rude for u to post another one?(woulda loved to hv a crack at ur question but. . .)

ooooooooooooooooo my my my i am soooooooooooooooooooooooo sorry bii i think it is infinity thanks metose.

ooooooooooooops i got ur name wrongly spelt sorry METOTSE

biina:

(Hard)
Find (as shown below) the sum of the infinite series, with the nth term defined as n*r^(n-1).

1 + 2r + 3r^2 + 4r^3 + 5r^4 + 6r^5 + 7r^6 +  . . .   . . . . . . . . . . = ?

where |r|<1

1. Infinity
2. 0
3. 1/(1-r)
4. r/(r-1)
5. 1/(1-r)^2
6. (r^2)/(1-r)^2
7. 1/(1-r^2)
8. ln(1-r)

Notes:
^ is the exponentiation function, and |x| is the absolute value of x.

Hint:
1. The nth term in the series above, n*r^(n-1), can be obtained by differentiating the nth term of the geometric progression (GP) series, r^n i.e. d(r^n)/dr = n*r^(n-1)
2. Differentiation is a linear operation i.e. dx+dy=d(x+y)
3. The sum of a GP series with first term a=1, and ratio r is easily found.

**Please no guessing. Answers should be followed by a brief comment explaining the reason for the choice**

The answer is 1/(1-r)^2

From the infinite sum of the GP series with first term a=1, and ratio r , we have

1 + r + r^2 + r^3 + r^4 + r^5 + r^6 +  . . .   . . . . . . . . . . = 1/(1-r)

Differentiating both sides with respect to r, we have


1 + 2r + 3r^2 + 4r^3 + 5r^4 + 6r^5 + 7r^6 +  . . .   . . . . . . . . . . = d[1/(1-r)]/dr=1/(1-r)^2

U people don pas my power ooooo

have u accepted my apology bii,,,,,,,,,,, and u too metotse

OMG(o my God) can u beleive it that its now i know itsNNNNNNNNNNNNNNNNetotse not MMMMMMMMMMetotse sorry dude. ;

biina , nice

2 Angels:

have u accepted my apology bii,,,,,,,,,,, and u too metotse

u don't have to apologize, as your actions weren't deliberate.
You can repost your earlier question, but remember to include options as the question are to be in multi-choice format.

thanks bii,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,'

a woman had 15naira spent 9 naira , ganed 12naira and loss 48 naira how much does she have left.
a) onaira
b) 8''''''''''
c) 9''''''''''''
d) 98'''''''''