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# Problem A. 546. (November 2011)

A. 546. Show that

for every positive integer k.

(5 pont)

Deadline expired on December 12, 2011.

Solution 1. Let Un(x) be the nth Chebyshev-polynomial of the second kind, for which Un(cos t)sin t=sin ((n+1)t). The roots of U2k are the numbers , , ..., and their negatives, xk+1=-x+1, xk+2=-x2, ..., x2k=-xk. On the left-hand side of the statement we have the sum of the squares of these numbers; the statement is equivalent with

 (1)

Lemma 1. U0(x)=1, U2(x)=4x2-1, and U2k+2(x)+U2k-2(x)=(4x2-2)U2k(x) for k1.

Proof. From U0(cos t)sin t=sin t, we get U0(x)=1. Similarly, from U2(cos t)sin t=sin (3t)=(4cos2-1)sin t we get U2(x)=4x2-1.

From the identity sin (a+b)+sin (a-b)=2sin acos b we obtain

U2k+2(cos t)sin t+U2k-2(cos t)sin t=

=sin ((2k+3)t)+sin ((2k-1)t)=2sin ((2k+1)t)cos (2t)=

=2U2k(cos t)sin t.(2cos2-1),

so U2k+2(x)+U2k-2(x)=(4x2-2)U2k(x).

Lemma 2. Let . Then ak,0=(-1)k, ak,1=0 és ak,2=(-1)k+1.2k(k+1).

Proof. Apply induction on k. If k=0 then U0(x)=1, so a0,0=1=(-1)k, a0,1=0, and a0,2=0=-4k(k+1) If k=1 then U0(x)=4x2-1, so a1,0=-1=(-1)k, a1,1=0, and a1,2=4=2k(k+1). Hence, the lemma holds for k=0 and k=1.

For the induction step assume the lemma for k and (k-1). By Lemma 1,

From this we read

ak+1,0=-2.(-1)k-(-1)k-1=(-1)k+1,

ak+1,1=0,

finally

ak+1,2=4.(-1)k-2.(-1)k+1.2k(k+1)-(-1)k.2k(k-1)=

=(-1)k(4+4k(k+1)-2k(k-1))=(-1)k+2.2(k+1)(k+2).

The lemma is proved.

Now from the Viéte formulae we find

Solution 2. Let , then egész számra .

From the identity we have

so

Applying again,

Therefore,

Solution 3. We use U2k again and prove (1). The polynomial can be factorized as , where c is the leading coefficient. Then

and

Substituting x=0 we obtain

 (2)

We compute the values U2k(0), U2k'(0), U2k''(0) by differentiating

 U2k(cos t)sin t=sin ((2k+1)t) (3a)

twice.

 -U2k'(cos t)sin2t+U2k(cos t)cos t=(2k+1)cos ((2k+1)t), (3b)
 U2k''(cos t)sin3t-3U2k'(cos t)sin tcos t-U2k(cos t)sin t=-(2k+1)2sin ((2k+1)t). (3c)

By substituting t=/2 in (3a--3c) we get U2k(0)=(-1)k and U2k'(0)=0, furthermore

U2k''(0)=(-1)k-(-1)k.(2k+1)2=(-1)k+1.4k(k+1).

Plugging these into (2),

### Statistics:

 7 students sent a solution. 5 points: Ágoston Tamás, Gyarmati Máté, Janzer Olivér, Kovács 444 Áron, Mester Márton, Omer Cerrahoglu, Szabó 928 Attila.

Problems in Mathematics of KöMaL, November 2011