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Maths Puzzle
(1!*1)+(2!*2)+(3!*3).......(n!*n)the sum to n terms of the above series is?
Read Solution (Total 3)
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- (1!*1)+(2!*2)+(3!*3).......(n!*n =(n+1)! -1
check, when
n=2,
(1!*1)+(2!*2)= 1+4=5 = 3!-1
when n=3,
(1!*1)+(2!*2)+(3!*3)= 1+4+18 = 23=4!-1
- 12 years agoHelpfull: Yes(2) No(0)
- x = (1!*1)+(2!*2)+(3!*3).......(n!*n)
y = 1! + 2! + 3!.......... n!
x + y = (1!*2)+(2!*3)+(3!*4).......(n!*n+1)
x + y = 2! + 3! + ...........(n+1)!
x + y + 1 = 1! + 2! + 3! +......(n+1)!
x + y + 1 = y + (n + 1)!
x + 1 = (n + 1)!
x = (n + 1)! - 1
enjoyyy...... - 12 years agoHelpfull: Yes(1) No(1)
- 0≤ k ≤n
==>∑k!*k
=∑[k!*{(k+1)-1}]
=∑[k!*(k+1)-k!]
=∑(k+1)!- ∑k!
=(n+1)!-1 - 11 years agoHelpfull: Yes(0) No(0)
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