Advanced book on Mathematics Olympiad

Real Analysis 497

=

m∑− 1

n= 0

[

( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn)−( 1 −ζ)( 1 −ζ^2 )···( 1 −ζn+^1 )

]

= 1 − 0 = 1 ,

and the identity is proved.

367.We have

1 +

∑r−^1

k= 0

(

r

k

)

Sk(n)= 1 +

∑r−^1

k= 0

(

r

k

)∑n

p= 1

pk= 1 +

∑n

p= 1

∑r−^1

k= 0

(

r

k

)

pk

= 1 +

∑n

p= 1

[(p+ 1 )r−pr]=(n+ 1 )r.

368.Setbn=

√

2 n−1 and observe that 4n=b^2 n+ 1 +b^2 n. Then

an=

bn^2 + 1 +b^2 n+bn+ 1 bn

bn+ 1 +bn

=

(bn+ 1 −bn)(b^2 n+ 1 +bn+ 1 bn+b^2 n− 1 )

(bn+ 1 −bn)(bn+ 1 +bn)

=

bn^3 + 1 −b^3 n

bn^2 + 1 −b^2 n

=

1

2

(bn^3 + 1 −bn^3 ).

So the sum under discussion telescopes as

a 1 +a 2 +···+a 40 =

1

2

(b^32 −b^31 )+

1

2

(b 33 −b^32 )+···+

1

2

(b^341 −b^340 )

=

1

2

(b^341 −b^31 )=

1

2

(

√

813 − 1 )= 364 ,

and we are done.
(Romanian Team Selection Test for the Junior Balkan Mathematical Olympiad, pro-
posed by T. Andreescu)

369.The important observation is that

(− 1 )k+^1

12 − 22 + 32 −···+(− 1 )k+^1 k^2

=

2

k(k+ 1 )

.

Indeed, this is true fork=1, and inductively, assuming it to be true fork=l, we obtain

12 − 22 + 32 −···+(− 1 )l+^1 l^2 =(− 1 )l+^1

l(l+ 1 )

2

.

Then