Summation Operator
 

Hi,
I'm trying to work out a way of getting Excel 2003 to answer

y=2i+1 when i=1 initially with a final value of 10.
y=(2x1+1)+(2x2+1)+...+(2x10+1)

I could type it by hand however my next question requires the upper limit n
to be infinity for the eqn

y=1/i(i-1) when i=1 initially and the final value for i=infinity
y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1)

and I eed to find the approximate answer for y when the addition of the
successive terms causes a lass than 0.0001% change in y, as well as the last
value for i when the program stops.

Am utterly stumped! Please help,

Thank you.

Summation Operator
 

In A1 enter:
..5
In A2 enter:
=A1+1/(ROW()*(ROW()+1)) and copy down
In B2 enter:
=(A2-A1)/A1 and copy down

For the first 10 rows:

0.50000000
0.66666667 33.3333333333%
0.75000000 12.5000000000%
0.80000000 6.6666666667%
0.83333333 4.1666666667%
0.85714286 2.8571428571%
0.87500000 2.0833333333%
0.88888889 1.5873015873%
0.90000000 1.2500000000%
0.90909091 1.0101010101%

and for rows 999 thru 1002:

0.99900000 0.0001002004%
0.99900100 0.0001000001%
0.99900200 0.0000998004%
0.99900299 0.0000996013%

so for i1000 the relative change term-to-term will be under your limit.
--
Gary''s Student - gsnu200797

Summation Operator
 

y=2i+1 when i=1 initially with a final value of 10.

y=(2x1+1)+(2x2+1)+...+(2x10+1)


Hi. For this first question...

n=10

? n + n*(1 + n)
120

For Second question:

Chg= 0.000001

? (Sqr(Chg^2 + Chg) - Chg)/Chg
999.000499999875

Hence, 999 is the last value.
1,000 will put you over.


and I need to find the approximate answer for y when the addition of the
successive terms causes a less than 0.0001% change in y


n=999

? n/(1 + n)
0.999

--
HTH :)
Dana DeLouis


"Sapphy" wrote in message ...

Hi,
I'm trying to work out a way of getting Excel 2003 to answer

y=2i+1 when i=1 initially with a final value of 10.
y=(2x1+1)+(2x2+1)+...+(2x10+1)

I could type it by hand however my next question requires the upper limit n
to be infinity for the eqn

y=1/i(i-1) when i=1 initially and the final value for i=infinity
y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1)

and I eed to find the approximate answer for y when the addition of the
successive terms causes a lass than 0.0001% change in y, as well as the last
value for i when the program stops.

Am utterly stumped! Please help,

Thank you.

Summation Operator
 

Hi. I posted a correction a while ago, but I still don't see it. Let me try again. I hope this isn't posted twice.

I made a mistake for #2.
When we round down, then the percentage difference between (n+1) and n has not reached our desired percentage change.
Hence, we need to Round Up.
If Chg equals 0.000001
Then

=CEILING((SQRT(Chg^2 + Chg) - Chg)/Chg,1)
Returns 1000.

With n = 1000, then this last value is
=n/(n+1)
=0.999000999000999


For the first question, when n=10, then:
=n*(2+n)
Returns 120.

--
HTH :)
Dana DeLouis


"Sapphy" wrote in message ...

Hi,
I'm trying to work out a way of getting Excel 2003 to answer

y=2i+1 when i=1 initially with a final value of 10.
y=(2x1+1)+(2x2+1)+...+(2x10+1)

I could type it by hand however my next question requires the upper limit n
to be infinity for the eqn

y=1/i(i-1) when i=1 initially and the final value for i=infinity
y=(1/(1(1+1))+(1/2(2+1)+...(1/infinity(infinity+1)

and I eed to find the approximate answer for y when the addition of the
successive terms causes a lass than 0.0001% change in y, as well as the last
value for i when the program stops.

Am utterly stumped! Please help,

Thank you.