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can anyone help with these sums?

 

Find 1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(99*100).

 

Find 1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + ... + 1/(98*99*100).

 Apr 30, 2020

Best Answer 

 #1
avatar+651 
+1

(1/1*2)+(1/2*3)+(1/3*4)+(1/4*5)+ .............+(1/99*100) =

= [1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... + [1/98 - 1/99] + [1/99 - 1/100] = 1/1 - 1/100 = 99/100 

 

:D

 Apr 30, 2020
 #1
avatar+651 
+1
Best Answer

(1/1*2)+(1/2*3)+(1/3*4)+(1/4*5)+ .............+(1/99*100) =

= [1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... + [1/98 - 1/99] + [1/99 - 1/100] = 1/1 - 1/100 = 99/100 

 

:D

LuckyDucky Apr 30, 2020
 #2
avatar+111435 
+1

Very nice, LuckyDucky!!!!

 

cool cool cool

CPhill  Apr 30, 2020
 #3
avatar+651 
+1

Thanks!

LuckyDucky  Apr 30, 2020
 #4
avatar
0

1 - sumfor(n, 1, 99, 1 / (n*(n+1)) = 99 / 100

 

2 - sumfor(n, 1, 98, 1 / (n^3 +3*n^2+ 2*n))= 4949 / 19800

 Apr 30, 2020

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