help solving this

[ibrahim]

x=(2^n)(1/20)^2(19/20)^n-2
n=4
find t for the following?
t=[FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]20*i*x[/FONT]

 

[tkhunny]

I cannot understand the notation. Please give it another go.

 

[ibrahim]

I cannot understand the notation. Please give it another go.

if x=(2^n)(1/20)^2(19/20)^(n-2)
n=4

t=[FONT=MathJax_Size1]∑ [[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Main]=1, [/FONT][FONT=MathJax_Main]∞][/FONT][FONT=MathJax_Math]20*i*x
what is the value of t?[/FONT]

 

[daon2]

if x=(2^n)(1/20)^2(19/20)^(n-2)
n=4

t=[FONT=MathJax_Size1]∑ [[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Main]=1, [/FONT][FONT=MathJax_Main]∞][/FONT][FONT=MathJax_Math]20*i*x
what is the value of t?[/FONT]

x nor 20 depend on i, so:

\(\displaystyle t = 20x\sum_{i=1}^{\infty} i\)

Obviously this sum does not converge.

 

[ibrahim]

Thank you daon2 v. much

x nor 20 depend on i, so:

\(\displaystyle t = 20x\sum_{i=1}^{\infty} i\)

Obviously this sum does not converge.

Thank you Sir