GMB 8 / 2005

 In GMB 8/2005, pages 378 (romanian version) and 380 (english version)

(other magazines here)

               "C : 2909.  For any positive integer  $n$  let  

$S_n=1\cdot 2\cdot 3+2\cdot 3\cdot 4+\dots +n(n+1)(n+2).$  Prove that :

a)  $\sqrt{S_n}\in\mathbb{R}\setminus \mathbb{Q}$ , for all  $n\in \mathbb{N^*}$  ;  b)  $\sqrt{\frac{1}{4}+S_n}\in\mathbb{Q}$ , for all  $n\in\mathbb{N^*}.$

{author :} Petre SIMION, Bucharest"

ANSWER CiP

$$S_n=\sum_{k=1}^nk\cdot (k+1) \cdot (k+2)=\fbox{$\frac{n(n+1)(n+2)(n+3)}{4}$} \tag{1}$$

a) It should be (cf. Lemma)   $\sqrt{S_n}\in\mathbb{N}$ , but  $\frac{n^2+3n}{2}<\sqrt{S_n}<\frac{n^2+3n}{2}+1$          (2)

b) $\sqrt{\frac{1}{4}+S_n}=\frac{n^2+3n+1}{2} \tag{3}$

Solution CiP , with COPILOT help

(for the help given by COPILOT see the draft at the end)

          We observe that  

$k(k+1)(k+2)=\frac{k(k+1)(k+2)\cdot 4}{4}=k(k+1)(k+2)\cdot \frac{(k+3)-(k-1)}{4}$ 

so

$k(k+1)(k+2)=\frac{(k+3)(k+2)(k+1)k}{4}-\frac{(k+2)(k+1)k(k-1)}{4} \tag{4}$

Summing in  (4)  from  $k=1\; to\;k=n$ , we obtain on the right the telescopic sum

$\left ( \color{Blue}{\frac{4\cdot 3\cdot 2\cdot 1}{4}}-\frac{3\cdot 2\cdot 1\cdot 0}{4}\right )+\left (\color{Orange}{\frac{5\cdot 4\cdot 3\cdot 2}{4}}-\color{Blue}{\frac{4\cdot 3\cdot 2\cdot 1}{4}}\right)+$

$+\left (\frac{6\cdot 5\cdot 4\cdot 3}{4}-\color{Orange}{\frac{5\cdot 4\cdot 3\cdot 2}{4}} \right )+\dots +\left (\frac{(n+3)(n+2)(n+1)n}{4}-\frac{(n+2)(n+1)n(n-1)}{4}\right)$

Terms of the same color are reduced and the result from  (1)  remains.

                    a)  $\sqrt{S_n}=\frac{\sqrt{n(n+1)(n+2)(n+3)}}{2}$  so according to Lemma , we must have

  $\sqrt{n(n+1)(n+2)(n+3)}\in\mathbb{N}\tag{5}$.

But the number in  (5)  is even because out of four consecutive numbers  $n\;,\;n+1\;,\;n+2\;,\;n+3\;$ , one is a multiple of  4 , so

$\frac{\sqrt{n(n+1)(n+2)(n+3)}}{2}\in\mathbb{N} \tag{5'}$

          But it is easy to show that we have the inequalities  (2)

 [    $\color{Red}{n(n+3)}=n^2+3n<\color{Red}{n^2+3n+2}\;\;\Leftrightarrow $$\Leftrightarrow\;\;\color{Red}{n(n+3)<(n+1)(n+2)} \tag{6}$

$\Rightarrow\; n^2(n+3)^2<n\color{Red}{(n+1)(n+2)}(n+3)\;\Rightarrow\;\frac{n(n+3)}{2}<\sqrt{S_n}$  so the left side in (2). It also results from  (6)  $\color{Red}n(n+1)(n+2)\color{Red}{(n+3)}<(n+1)^2(n+2)^2\;\Rightarrow\;\sqrt{S_n}<\frac{(n+1)(n+2)}{2}=\frac{n^2+3n}{2}+1$  so the right side in (2). Note that  $\frac{n^2+3n}{2}\in\mathbb{N}\;(\forall) n\in\mathbb{N}.$      ]

Then  (5)  is impossible, because it is strictly contained between two consecutive integers.

                     b)  We can write  $S_n=\frac{n(n+3)(n+1)(n+2)}{4}=\frac{(n^2+3n)(n^2+3n+2)}{4}=$

$=\frac{[(n^2+3n+1)-1][n^2+3n+1)+1]}{4}=\frac{(n^2+3n+1)^2-1}{4}$  so  $\sqrt{\frac{1}{4}+S_n}$  has the value from  (3).

$\blacksquare$

          Remark CiP  Draft of the discussion with COPILOT :

<<

Suma

[ S = 1\cdot 2\cdot 3 ;+; 2\cdot 3\cdot 4 ;+; \dots ;+; n(n+1)(n+2) ]

are o expresie închisă foarte elegantă:

[ \boxed{S = \frac{n(n+1)(n+2)(n+3)}{4}} ]

---

De ce funcționează formula

Termenul general este:

[ k(k+1)(k+2) = k^3 + 3k^2 + 2k ]

Suma devine:

[ S = \sum_{k=1}^n (k^3 + 3k^2 + 2k) ]

Folosind formulele standard pentru
(suma;cuburilor),
(suma;pătratelor),
(suma;primelor;n;numere),

și simplificând, totul se reduce exact la:

[ S = \frac{n(n+1)(n+2)(n+3)}{4} ]

---

Verificare rapidă pentru n = 1, 2, 3

• n = 1:
  Stânga: (1\cdot 2\cdot 3 = 6)
  Dreapta: (\frac{1\cdot 2\cdot 3\cdot 4}{4} = 6)

• n = 2:
  Stânga: (6 + 24 = 30)
  Dreapta: (\frac{2\cdot 3\cdot 4\cdot 5}{4} = 30)

• n = 3:
  Stânga: (6 + 24 + 60 = 90)
  Dreapta: (\frac{3\cdot 4\cdot 5\cdot 6}{4} = 90)

Totul se potrivește perfect.

---

Dacă vrei, pot și demonstra formula prin:

• inducție
• metoda diferențelor finite
• identități telescopice

Vrei o demonstrație completă printr-una dintre aceste metode?

Telescopica

Hai să o facem telescopic, „curat”, cu combinații.

---

1. Scriem termenul general în formă combinatorică

Observăm că

[ k(k+1)(k+2) = 6 \cdot \binom{k+2}{3} ]

pentru că

[ \binom{k+2}{3} = \frac{(k+2)(k+1)k}{6}. ]

Deci suma devine

[ S = \sum_{k=1}^n k(k+1)(k+2) = 6 \sum_{k=1}^n \binom{k+2}{3}. ]

---

2. Transformăm suma într-o sumă telescopică

Folosim identitatea clasică:

[ \binom{m}{r} = \binom{m+1}{r} - \binom{m+1}{r-1}. ]

Aplicată pentru (r = 3), obținem:

[ \binom{k+2}{3} = \binom{k+3}{4} - \binom{k+2}{4}. ]

Deci

[ \sum_{k=1}^n \binom{k+2}{3} = \sum_{k=1}^n \left[\binom{k+3}{4} - \binom{k+2}{4}\right]. ]

Aceasta este o sumă telescopică: termenii se anulează în lanț.

Scriem câțiva termeni:

[ \begin{aligned} &\left[\binom{4}{4} - \binom{3}{4}\right] + \left[\binom{5}{4} - \binom{4}{4}\right] + \left[\binom{6}{4} - \binom{5}{4}\right] + \dots + \left[\binom{n+3}{4} - \binom{n+2}{4}\right]. \end{aligned} ]

Tot ce e „intermediar” se taie, rămâne doar:

[ \sum_{k=1}^n \binom{k+2}{3} = \binom{n+3}{4} - \binom{3}{4}. ]

Dar (\binom{3}{4} = 0), deci

[ \sum_{k=1}^n \binom{k+2}{3} = \binom{n+3}{4}. ]

---

3. Punem totul la un loc

Revenim la suma inițială:

[ S = 6 \sum_{k=1}^n \binom{k+2}{3} = 6 \binom{n+3}{4}. ]

Acum calculăm:

[ 6 \binom{n+3}{4} = 6 \cdot \frac{(n+3)(n+2)(n+1)n}{4!} = 6 \cdot \frac{(n+3)(n+2)(n+1)n}{24} = \frac{n(n+1)(n+2)(n+3)}{4}. ]

---

Concluzie

[ \boxed{ \sum_{k=1}^n k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4} } ]

Dacă vrei, putem face și o versiune pur algebrică de telescopare, găsind direct un polinom (F(k)) cu (F(k+1)-F(k)=k(k+1)(k+2)).

>>

{end Rem}

A cartoon of the Stolz–Cesàro Theorem // La caricatura di Teorema Stolz-Cesàro

 In the Mathematical Review of Students from Timișoara 1/1978, RMT for short, I found a problem (page 69) that I solved in my time :

 "3304.  Given the sequences  $\{a_n\}_{n\in\mathbb{N}}\;,\;\;\{b_n\}_{n\in\mathbb{N}}$ with the properties :

 (i)  $\{b_n\}_{n\in\mathbb{N}}$  is strictly monotone and unbounded.

                               (ii)  exists  $\displaystyle \lim_{n\to \infty}\frac{a_n}{b_n}$

                               (iii)  $\frac{a_{n+1}}{a_n}+\frac{b_{n+1}}{b_n}=2\;,\;\;(\forall) n\in\mathbb{N}.$

                           Prove that  $\displaystyle \lim_{n \to \infty}\frac{a_n}{b_n}=0.$

{author :} Titu ANDREESCU, student, Timișoara"

Solution CiP

              The condition  (iii)  is written equivalently : $\frac{a_{n+1}}{a_n}-1+\frac{b_{n+1}}{b_n}-1=0\;\Leftrightarrow$

$\Leftrightarrow\;\frac{a_{n+1}-a_n}{a_n}=-\frac{b_{n+1}-b_n}{b_n}\;\Leftrightarrow\;\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=-\frac{a_n}{b_n} \tag{1}$

By  (ii)  the limit  $\displaystyle\lim_{n\to \infty}\frac{a_n}{b_n}$  exists, and then from  (1)  it follows that the limit  $\displaystyle \lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$  also exists. Then, according to Stolz–Cesàro theorem

$\displaystyle \lim_{n\to \infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\overset{(1)}{=}-\lim_{n\to\infty}\frac{a_n}{b_n}$,

hence  $\displaystyle \lim_{n\to \infty}\frac{a_n}{b_n}=0.$

QED

$\blacksquare$

A necessary correction: it seems that I was wrong in my calculations, not only in life

 The answer to the Problem in the post here is WRONG. It was brought to my attention by a commenter, unfortunately anonymous.

          The answer will be replaced there with this one :

Corrected   ANSWER CiP

The statement is NOT true for polynomials of degree 4 :

If  $P(x)=x^4\;,\;\;Q(x)=x^4+x^2\;,\;\; R(x)=2x^4+x^2\;\;$  then  $P(x)\leqslant Q(x)\leqslant R(x)$

but  $\lambda \cdot P(x)+(1-\lambda)\cdot R(x)=(2-\lambda)\cdot x^4+(1-\lambda)\cdot x^2\neq Q(x)$

$\square$

Three Nice Polynomials of Degree III // Tre pæne polynomier af grad III

 We will solve Problem 7 (page 38, in the magazine from yesterday's Post) from Team Selection Test(TSTs) 1 for IMO.

" Problem 7.  Consider three polynomials of degree 3 with real 

coefficients $P,\;Q,\;R$  such that  $P(x)\leqslant Q(x)\leqslant R(x)$  for any real  $x$

  and, in addition, there exists a real number  $a$  such that  $P(a)=R(a).$

  Show that there exists a constant  $\lambda \in [0,\;1]$  with the property

  $Q=\lambda P+(1-\lambda)R.$  Does the statement remain true in the case 

                         when  $P,\;Q,\;R$  have degree 4?

{authors : } I. Cuculescu and L. Panaitopol, Bucharest"

ANSWER CiP (see correction)

It is not entirely true for polynomials of degree 4. For example, if 

$P(x)=x^4\;,\;\;Q(x)=x^4+x^2\;,\;\;R(x)=x^4+2x^2$  then  $P(x)\leqslant Q(x)\leqslant R(x)$

  and we have $Q(x)=\lambda \cdot P(x)+(1-\lambda)\cdot R(x)$  but with  $\lambda =-1 \not \in [0,\;1]$

Solution CiP

                   Obviously  $P(a)=Q(a)=R(a).$  Let's define polynomials

$S(x)=R(x)-Q(x)\;\geqslant 0\;,\;\forall x\;\;;\;\;T(x)=R(x)-P(x)\;\geqslant 0\;,\;\forall x \tag{1}$

          We have  $S(a)=0$  therefore  $S(x)=(x-a)S_2(x)$ , for a certain polynomial  $S_2$  of degree 2.

Since when  $x$  passes through the value  $x=a$ , the polynomial  $S$  does not change sign, we must have 

$S(x)=(x-a)^2\cdot S_1(x) \tag{2}$

with  $S_1$-a first degree polynomial. But the polynomial  $S_1$  should have a root, through which if   $x$  passes, the expression in  (2)  changes sign again. Contradiction, so  $S_1$  is a constant.

          In exactly the same way results

$T(x)=(x-a)^2\cdot T_1(x) \tag{3}$

with  $T_1$  a constant.

          Let  $\lambda :=\frac{S(x)}{T(x)}\overset{(2)}{\underset{(3)}{=}}\frac{S_1}{T_1}=constant\;$.  Since  $\lambda=\frac{R(x)-Q(x)}{R(x)-P(x)}$  and  $R(x)-Q(x)\leqslant R(x)-P(x)$  we have  $0\leqslant \lambda \leqslant 1$  and 

 $Q(x)=\lambda \cdot P(x)+(1-\lambda )\cdot R(x).$

$\blacksquare$

Problem E : 6271

 Click on the image and use the password : ogeometrie . The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.

On page 292 :

                          " E : 6271*.  Find the maximum and minimum of the sum  $x+y$

                                                   and the product  $xy$ , knowing that  $5x+6y=150$ ,

 and  $x\; and\; y$  are natural numbers different from zero.           

{author : } Gh. MARGHESCU, student, București "

ANSWER CiP

$max\;(x+y)=29$  for  $x=24\;, \;y=5$

$min\;(x+y)=26$   for  $x=6\;,\;y=20$

$max\;(x\cdot y)=180$  for  $x=12\;,\;y=15\;\;or\;\;x=18\;,\;y=10$

$min\;(x\cdot y)=120$  for  $x=6\;,\;y=20\;\;or\;\;x=24\;,\;y=5$

Solution CiP

                  We have the relationships :

$5x+6y=150\;\;,\;\;\;5x=150-6y\;\;\;6y=150-5x \tag{1}$

          From  $6\mid 150\;\;and\;\;6\mid 6y$  and from the second formula (1) it follows that  $6\mid 5x$  and because  $gcd(5,6)=1$  we have  $6\mid x$  so

$x=6\cdot x_1\;,\;\;x_1\in\mathbb{N^*} \tag{2}$

Similarly, we have  $5\mid 150\;\;and\;\;5\mid 5x$ , and from the third formula (1) it results  $5\mid 6y$ , so

$y=5\cdot y_1\;\;\;y_1\in\mathbb{N^*} \tag{3}$

          Then  $5x+6y=150\;\;\overset{(2)}{\underset{(3)}{\Leftrightarrow}}\;\;\;5\cdot 6x_1+6\cdot 5y_1=150\Leftrightarrow$

$\Leftrightarrow\;\;\;\;\;\;\;\;\;\;x_1+y_1=5 \tag{4}$

If  $x_1=n\in\mathbb{N^*}$ , then  $y_1\overset{(4)}{=}5-x_1=5-n$ , so the nonzero natural number solutions for  (4)  are :

$x_1=n\;\;,\;\;\;y_1=5-n\;\;\;\;\;1\leqslant n \leqslant 4 \tag{5}$

But

$x+y\;\;\;\overset{(2)\;(3)}{\underset{(5)}{=}}\;\;6\cdot n+5\cdot (5-n)=n+25 \tag{6}$

and then  $1\leqslant n \leqslant 4 \Leftrightarrow26\leqslant n+25 \leqslant 29\underset{(6)}{\Leftrightarrow}$

$\Leftrightarrow\;\;\;\;\;26 \leqslant x+y \leqslant 29 \tag{7}$

We have in  (7)  $x+y=26$  for  $n=1$ , so  $x_1=1\;,\;y_1=4$  and we get from  (2)  and  (3)  $x=6\;,\;y=20$  for the minimum value.  And  $x+y=29$  for  $n=4$ , so $x_1=4\;,\;y_1=1$  so  $x=24\;,\;y=5$  for the maximum value.

          Since there are a small number of values ​​for  $x\; and\; y$ , we will list all the products  $x\cdot y$  in the table below, from where we will also obtain the answer.

\begin{array}{c|c|c|c|c}n&1&2&3&4\\\hline x&6&12&18&24\\\hline y&20&15&10&5\\\hline x\cdot y&120&180&180&120\\\end{array}

$\blacksquare$

The E : 17461 problem, half easy, half hard // Задача E: 17461, наполовину простая, наполовину сложная

 From GMB (page 159), also proposed for 5th grade.

          " E : 17461.  Let  $N=3+3^2+3^3+\dots+3^{4n+1}\;,\;n\in\mathbb{N}.$

            a)  Show that  $2N+3$  s a perfect square.

            b)  For  $n$  even natural numbers, determine the last two digits of  $N$.

                                                                                  {author : } Marin CHIRCIU, Pitești "

Partial  ANSWER  CiP

a)  $2N+3=(3^{2n+1})^2$

b) The answer is not a unique number. We have :

$3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9=29\;5\color{Yellow}{23}$

$3+3^2+3^3+\dots+3^{17}=193\;710\;2\color{Pink}{43}$

Solution CiP

                    a) $2N+3=\color{Red}{3}+N+N=$

$=\color{Red}{3}+(3+3)+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1})=$

$=\color{Red}{(3+3+3)}+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1}).$

Let's observe that  $3+3+3=3^2$ ...and in general

$3^k+3^k+3^k=3^{k+1} \tag{1}$

Then, the red parenthesis results is $\color{Red}{3^2}$ , which appears as the term in the second parenthesis thus :

$(\color{Red}{3^2}+3^2+3^2)+(3^3+3^3)+\dots +(3^{4n+1}+3^{4n+1}).$

     Continuing this calculation, we finally obtain the result

$2N+3=\color{Red}{3^{4n+1}}+3^{4n+1}+3^{4n+1}\underset{(1)}{=}3^{4n+2}=(3^{2n+1})^2\;,$

i.e. a perfect square.

$\square$

Our Beloved Magazine "GAZETA MATEMATICĂ series B" - GMB 3 / 2026 // Notre Magazine Bien-Aimé "GAZETA MATEMATICĂ série B" - GMB 3 / 2026

  Click on the image and use the password : ogeometrie .The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.

          A problem to illustrate the love for numbers, proposed for 5th grade (page 159) :

                    "E : 17457 .   Write the number  $220^{2n+1}$  as the sum of 5 distinct

                      nonzero perfect squares. 

{author : } Marin CHIRCIU, Pitești "

ANSWER CiP

$220=1^2+3^2+4^2+5^2+13^2$  hence

$220^{2n+1}=(220^n)^2+(3\cdot 220^n)^2+(4 ^2\cdot 220^n)^2+(5\cdot 220^n)^2+(13\cdot 220^n)^2$

Solution CiP

               We have  $220^{2n+1}=(220^n)^2\cdot 220^1$ , so let's try to write the number  $220$  as the sum of five distinct nonzero perfect squares. 

We write some sums,  $\color{Green}{highlighting}$  a perfect square term:

$220=\color{Green}{196}+24$

$220=\color{Green}{169}+51$

$220=\color{Green}{144}+76$

..........................

But any number is written as the sum of four squares, according to Lagrange's Four-Square Theorem, so let's try this with the second term in the writings above.

          Asking a friend I found out that : 

<< Adevărul e că  $24$ are foarte puține reprezentări valide.

Singura reprezentare cu pătrate întregi ne-negative este:

$$\boxed{{\displaystyle 24=4^2+2^2+2^2+0^2}}$$

>>

In translation : <<The truth is that 24 has very few valid representations. The only representation with non-negative integer squares is: ... >> But the squares here are not all nonzero and not all distinct.

Moving on to term $51$ , the same friend gave me the answer you see in the first row above.

The problem is now solved.

$\blacksquare$