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Find the sum of the series 1/1.2.3.4 + 1/2.3.4.5 +...... n terms.

5 years ago

Answers : (2)

vsujji
17 Points
							
nth term of the given series is 1/[n][n+1][n+2][n+3]
T={1/[n][n+1]}{1/[n+2][n+3]}
    ={1/n-1/[n+1] } {1/[n+2]-1/[n+3]}
    ={1/n}{1/[n+2]}-{1/[n]}{1/[n+3]}-{1/[n+1]}{1/[n+2]}+{1/[n+1]}{1/[n+3]}
    =1/2{1/[n]-1/[n+2]}-1/3{1/n-1/[n+3]}-[1/[n+1]-1/[n+2]}+1/2[1/[n+1]-1/[n+3]}
  …........
Tn =1/6{1/n}-1/2[1/[n+2]} +1/2{1/[n+2]}-1/6{[1/n+3]} 
         n=n
sum=∑Tn
        n=1 
5 years ago
grenade
2061 Points
							
telescopic series
5 years ago
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