Begin2.DVI

I11-11. (a)

(p+q)n=

(

n

0

)

pn+

(

n

1

)

pn−^1 q+···+

(

n

n−x

)

pxqn−x+···

(

n

n

)

qn

(

n

k

)

=

(

n

n−k

)

so that (p+q)n=

∑n

x=0

(

n

n−x

)

pxqn−x=

∑n

x=0

(

n

x

)

pxqn−x= 1 =

∑n

x=0

f(x)

(b)

(p+q)n−^1 =

n∑− 1

x=0

(

n− 1

n− 1 −x

)

pxqn−^1 −x

shift summation index by lettingx=X− 1 so that

(p+q)n−^1 =

∑n

X=1

(

n− 1

n− 1 −X+ 1

)

pX−^1 qn−X=

∑n

X=1

(

n− 1

X− 1

)

pX−^1 qn−X= 1

since

(

n− 1

n− 1 −X+ 1

)

=

(

n− 1

X− 1

)

(c)x

(

n

x

)

=x n!

x!(n−x)!

= n(n−1)!

(x−1)!(n− 1 −(x−1))!

=n

(

n− 1

x− 1

)

(d)

μ=

∑n

x=0

xf(x) =

∑n

x=1

x

(

n

x

)

pxqn−x=

∑n

x=1

n

(

n− 1

x− 1

)

pxqn−x=np

I11-12.

s^2 =^1

N− 1

∑N

j=1

(xj− ̄x)^2 =^1

N− 1

∑N

j=1

(x^2 j−2 ̄xxj+ ̄x^2 )

=^1

N− 1





∑N

j=1

x^2 j−2 ̄x

∑N

j=1

xj+N ̄x^2





=^1

N− 1





∑N

j=1

x^2 j−2 ̄xNx ̄+N ̄x^2



=^1

N− 1





∑N

j=1

x^2 j−Nx ̄^2





Use the fact thatx ̄=^1

N

∑N

j=1

xjand write

s^2 =

1

N(N−1)





N

∑N

j=1

x^2 j−





∑N

j=1

xj





2 



Solutions Chapter 11