Question

2!/1! +3!/2!+...... +(n+1)!/n! = n*(n+3)/2
n€N /0
Va rog frm
Dau coroana

Answer 1

2!/1!=1×2/1=2
3!/2!=1×2×3/1×2=3
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(n+1)!/n!=1×2×3×...×(n+1)/1×2×3×...×n=n+1

S = 2!/1!+3!/2!+...+(n+1)!/n!=2+3+...+n+1

(n+1)+2=n+3
n+3=n+3
(n-1)+4=n+3
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2+(n+1)=n+3

(n+1)-2+1=n
~n termeni de (n+3)

2 S = (n+3)×n

S = (n+3)×n/2