View Single Post
  #3   Report Post  
Posted to microsoft.public.excel.worksheet.functions
Dana DeLouis Dana DeLouis is offline
external usenet poster
 
Posts: 947
Default 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.