Answer:
k=3(n+1)(n+2)
Step-by-step explanation:
Given P(Ei)=in and let P(Ei)=ki(i+1)
so
∑i=1nP(Ei)=k∑i=1ni(i+1)=k[n(n+1)(2n+1)6+n(n+1)2]
1=kn(n+1)2[2n+13+1]=k3⋅(n+1)(n+2)
so k=3(n+1)(n+2)
