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(#M40000116) IIT-JEE 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 (16 years ago)

Solved Read Solution (1)

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(#M40026377) IIT-JEE QUESTION Keep an EYE

A client is celebrating his 50th b-day and wants to start saving for his anticipated retirement at age 65. he wants to be able to withdraw $20,000 from his saving account on his 66th b-day and each year for 19 more years after that.

After extensive research, the client determines that he can invest his money in an account that offers 5% interest per year with quarterly compounding. he wants to make equal annual payments on each b-day into the account - the first payment on his 51st b-day and his last on his 65th b-day.

In addition, the clients employer will contribute $2000 to the account each year (beginning on the clients 51st b-day) as part of the companies profit-sharing plan (a total of 15 contributions.)

The amount the client must deposit personally into the account each year on his b-day to enable him to make the desired wds at the retirement is closest to:

A. 9375

B. 9459

C. 11400

Asked In IIT-JEE Garima (12 years ago)

Solved chaitanya Read Solution (2)

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