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1(1!)+2(2!)+3(3!)+--------+2012(2012!)=?

Read Solution (Total 2)

1(1!)+2(2!)=3!-1
1(1!)+2(2!)+3(3!)=4!-1
1(1!)+2(2!)+3(3!)+4(4!)=5!-1
.....
.....
similarly,
1(1!)+2(2!)+3(3!)+.....+2012(2012!)=2013!-1
11 years agoHelpfull: Yes(99) No(2)

1(1!)+2(2!)+......+2012(2012!)=2013!-1
11 years agoHelpfull: Yes(1) No(0)

TCS Other Question

One card is lost out of 52 cards. two cards are drawn randomly. They are spade. What is the probability that the lost card is also spade? 1*2+2*2^2+3*2^3+4*2^4........+2012*2^2012 find the sum of the series ?