498 Real Analysis
12 − 22 + 32 −···+(− 1 )l+^2 (l+ 1 )^2 =(− 1 )l+^1
l(l+ 1 )
2+(− 1 )l+^2 (l+ 1 )^2=(− 1 )l+^2 (l+ 1 )(
−
l
2+l+ 1)
,
whence
(− 1 )l+^2
12 − 22 + 33 −···+(− 1 )l+^2 (l+ 1 )^2=
2
(l+ 1 )(l+ 2 ),
as desired. Hence the given sum equals
∑nk= 12
k(k+ 1 )= 2
∑nk= 1(
1
k−
1
k+ 1)
,
telescoping to
2
(
1 −
1
n+ 1)
=
2 n
n+ 1.
(T. Andreescu)370.The sum telescopes once we rewrite the general term as
1
(√
n+√
n+ 1 )(^4√
n+^4√
n+ 1 )=
√ (^4) n+ 1 −√ (^4) n
(
√
n+ 1 +√
n)(^4√
n+ 1 +^4√
n)(^4√
n+ 1 −^4√
n)=
√ (^4) n+ 1 −√ (^4) n
(
√
n+ 1 +√
n)(√
n+ 1 −√
n)=
√ (^4) n+ 1 −√ (^4) n
n+ 1 −n
=^4
√
n+ 1 −^4√
n.The sum from the statement is therefore equal to^4
√
10000 − 1 = 10 − 1 =9.
(Mathematical Reflections, proposed by T. Andreescu)371.As usual, the difficulty lies in finding the “antiderivative’’ of the general term.
We have
1
√
1 +( 1 +^1 n)^2 +√
1 +( 1 −^1 n)^2=
√
1 +( 1 +^1 n)^2 −√
1 +( 1 −^1 n)^2
1 +( 1 +^1 n)^2 − 1 −( 1 −^1 n)^2=
√
1 +( 1 +^1 n)^2 −√
1 +( 1 −^1 n)^2
4
n