Latest IIT-JEE Aptitude Question SOLUTION: sum to n terms: 0!/5! + 1!/6! + 2!/7! + ... ?

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(#M40000116) IIT-JEE Numerical Ability question sum to n terms Keep an EYE

sum to n terms: 0!/5! + 1!/6! + 2!/7! + ... ?

Solution

r th term of the series is (r-1)!/(r+4)!= 1/r(r+1)(r+2)(r+3)(r+4)
= (1/24r)-(1/6(r+1))+(1/4(r+2))-(1/6(r+3))+(1/24(r+4))
Now put r = 1,2,3,4,5.......upto n and add all the terms [here a special arrangement is that the solution of 1/24-1/6+1/4-1/6+1/24 is 0 so sum of terms from 1/5 to 1/n would be 0]
finally we will get
n(n+5)(n^2+5n+10)/96(n+1)(n+2)(n+3)(n+4)

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Asked In IIT-JEE (14 years ago)

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midhun prakash  (13 years ago)

3!/8!

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