1= \sum_{1}^{n}{\frac{1}{k!(k+2)}}
agey soch rahe hai.....
I saw something like this in Mathematics today sometime..
\sum_{1}^{n}{\frac{k+2}{k!+(k+1)!+(k+2)!}}
Find this....
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15 Answers
24haan sky yahaan tak to ho hi gaya........
Σ(k+2)/[k!+(k+1)k!+(k+2)(k+1)k!]
=>Σ(k+2)/k!(k+2)2
=>Σ1/k!(k+2)
13=Σk/(k+2)!+Σ1/(k+2)!
=Σ1/(k+1)! - Σ2/(k+2)! + Σ1/(k+2)!
24how about multiplying and dividing by k+1
so that it becomes Σ(k+1)/(k+2)!................dont know whether it will help or not....[12]
1direct substitution after the processing of the question into the simpler form........ sunil u'll get sky's answer...........
1after post no #2 :
= Σ(k=1 to n) [(k+1) / (k+2)!]
= Σ(k=1 to n) [2/3! + 3/4! + 4/5! + ........ + n/(n+1)!]
= Σ(k=1 to n) [(3-1)/3! + (4-1)/4! + ....... + {(n+1)-n}/(n+1)!]
= Σ(k=1 to n) [1/2! - 1/3! + 1/3! - 1/4! .............. + 1/n! - 1/(n+1)!]
= 1/2! - 1/(n+1)!
13\sum_{1}^{n}{\frac{k+1}{(k+2)!}}
=>\sum_{1}^{n}{\frac{k+2-1}{(k+2)!}}
\Rightarrow \sum_{1}^{n}{\frac{1}{(k+1)!}}-\sum_{1}^{n}{\frac{1}{(k+2)!}}
t1= 1/2! - 1/3!
t2= 1/3! - 1/4!
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tn=1/(n+1)!-1/(n+2)!
Sn=1/2! - 1/(n+2)!
small mistake in sky's solution....