solve aH1L=54, aHnL=-34 aHn-1L for aHnL
Input interpretation :
aH1L ‡
solve
5
4
aHnL ‡ -
3
4
for
aHnL
aHn - 1L
Result :
aHnL ‡ - 5 -
1
4
n
3n-1
Generated
by Wolfram|Alpha
(www.wolframalpha.com)
onJuly23, 2011 fromChampaign,
IL.
© WolframAlphaLLC—A Wolfram Research Company
1
sum -5 H-14L^n 3^H-1 + nL, n=1..infinity
Infinite sum :
¥
â- 5 n =1
1
n
4
3 -1+n ‡
5
7
» 0.714286
Partial sums :
1.2
æ
æ
1.0
æ
0.8
0.6
æ
æ
0.4
æ
0.2
2
3
4
5
6
Convergencetests:
By the alternating series test , the series converges .
Partial sum formula :
m
1 n -1+n
â- 5 n =1
4
3
‡-
5
7
-
3
4
m
-1
Generated
by Wolfram|Alpha
(www.wolframalpha.com)
onJuly23, 2011 fromChampaign,
IL.
© WolframAlphaLLC—A Wolfram Research Company
1
sum -5 (-1/4)^n 3^(-1 + n), n=1..infinity - Wolfram|Alpha
1 of 1
http://www.wolframalpha.com/input/?i=sum++-5+(-1/4)^n+3^(-1+++n)...
x
Infinite sum:
More digits
Partial sums:
More terms
Convergence tests:
Partial sum formula:
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23.7.2011 19:42