The document discusses the benefits of exercise for both physical and mental health. Regular exercise can improve cardiovascular health, reduce symptoms of depression and anxiety, enhance mood, and boost brain function. Staying physically active for at least 30 minutes each day is recommended for significant health advantages.
This document provides an overview and description of a book titled "Exercises in Modern English Grammar" by A. S. Saakyan. It was published in 2006 by Iris-Press as part of their higher education series. The book contains 448 pages and is intended as a study guide for high school students and university students to review and strengthen their knowledge of English grammar. It covers morphology and syntax through analytical, practice, and creative exercises. Each section concludes with test exercises and answers to allow for self-assessment. The book aims to reflect the latest developments in the English language from the late 20th and early 21st centuries. It is designed to be a flexible resource that can be used alongside any basic English language course.
Românismul de la Mihai Eminescu la Grigore Vieruinachirilov
Proiect “Educație online fără hotare” 2023 - 2024,
implementat de Direcția Generală Educație, Tineret și Sport a municipiului Chișinău în cadrul Proiectului “Educație online”
1. x
Calculul unor sume in gimnaziu
Exercitii in care se cere calcularea unei sume de mai multi termeni sunt intalnite
chiar in manualele de clasa a-IV-a sau a-V-a.Am considerat necesara demonstrarea
unor formule de calcul pentru acestea ,altele decat cele ce folosesc inductia
matematica sau o pseudo-inductie matematica,in ideea de a le folosi in rezolvarea
unor probleme propuse pentru diferite concursuri.
Calculul unor sume de numere
1. S= 1 +2 +3 + …+(n-2) +(n-1) +n
S=n +(n-1)+(n-2)+… + 3 +2 +1
2S=n+1+n+1+n+1+…+n+1+n+1+n+1
2S=n(n+1)
n( n +1)
S=
2
2. S=1 + 3 + 5 +…..+(2n-5)+(2n-3)+(2n-1)
S=(2n-1)+(2n-3)+(2n-5)+…+ 5 + 3 + 1
2S=2n + 2n +2n +…+ 2n + 2n + 2n
2S=2n.n
2
S= n
2 n −2 x n −1 n
3. S=1 + x + x +…+ x + x + x
x + x2 + x
3 n −1 n
Sx= + ... + . x +x
n +1
Sx-S = x −1
n +1
S(x-1) = x −1
n +1
S=( x -1)/( x -1)
2 2 2 2
4. S=
1 2 + 3 +…+ n
+
Folosind suma primelor n numere naturale impare putem scrie:
2
1 =1
1
2. 2
2 =1+3
2
3 =1+3+5
…………………………….
2
k =1+3+5+…+(2k-1)
…………;…………………..
2
n =1+3+5+…+(2k-1)+…+(2n-1)
Adunand membru cu membru obtinem:
S=n.1+(n-1).3+(n-2).5+…+(n-k+1).(2k-1)+…+2.(2n-3)+(2n-1)
Termenul general are forma:(2k-1).(n-k+1) si poate fi scris:
2
k +k,atunci:
(2k-1).(n-k+1)=(n+1).(2k-1)-2
2 2 2 2
S=(n+1).(1+3+5+…+2n-1)-2(
1 + 2 + 3 +…+ n )+(1+2+3+…+n)
2
3S=(n+1). n +n(n+1)/2
2
6S=2.(n+1). n +n.(n+1)
6S=n(n+1)(2n+1)
n(n +1)(2n +1)
S=
6
1 1 1 1
5. S= + + +…+ n( n +1)
1.2 2.3 3.4
1 1 1
Se demonstreaza usor ca: n( n +1) = - ⇒
n n +1
1 1 1 1 1 1 1 1 n
S= - + - +…+ - = - =
1 2 2 3 n n +1 1 n +1 n +1
k 1 1
Generalizare: n(n + k ) = -
n n +k
Aplicatii:
a) Calculati suma cifrelor numarului:
x=9+99+999+…+99..99,unde ultimul termen are 2008 cifre.
Numarul x se mai poate scrie:
2 3 2008 2 3 2008
x=10-1+10 -1+10 -1+…+10 -1=(10+10 +10 +…+10 -1=
2 3 2008 2 2007
=(10+10 +10 +…+10 )-2008=10(1+10+10 +…+10 )-2008=
2008
=10. 10
−1 999..99
-2008=10. -2008=10.111…11-2008=111…1109102.In
9
10 − 1
rezultat apare de 2004 ori,deci suma cifrelor va fi :2016.
Generalizare:
Pentru a calcula: S=a+ aa + aaa +…+ aa...aa se calculeaza:
2
3. a
(9+99+999+…+99…9)
9
3 5 7 85
b)Calculati: S= + + +…+
1.4 4.9 9.16 1764.1849
k 1 1
Se foloseste relatia: n(n + k ) = - si avem:
n n +k
1 1 1 1 1 1 1 1 1848
S= - + - + - +…+ - =
1 4 4 9 9 16 1764 1849 1849
c)Sa se calculeze:
1 1 1 1
S= 1.( k +1) + ( k +1)(2k +1) + (2k +1)(3k +1) +…+ [(n −)k +1](nk +1)
Se observa ca diferenta dintre factorii de la numitor este k,deci vom inmulti cu k si
obtinem:
k k k k
Sk= 1.( k +1) + (k +1)(2k +1) + (2k +1)(3k +1) +…+ [(n −1) k +1](nk +1) =
1 1 1 1 1 1 1 1
= - + - + - +…+ ( n −1) k +1 - =
1 k +1 k +1 2k +1 2k +1 3k +1 nk +1
1 1 nk +1 −1 nk n
= - = = ,de unde:S= .
1 nk +1 nk +1 nk +1 nk +1
d)Aratati ca numarul :
2 3 2006
N=1+2+
2 +2 2 nu este patrat perfect.
+…+
2007
Calculand N obtinem: N=
2 -1
2007 2007
U(
2 -1)=U(U( 2 )-1)=7.Cum nici un patrat perfect nu se termina in 2,3,7,8
rezulta N nu este patrat perfect.
e)Sa se calculeze suma:
(2n − )
2 2 2
1
2
S=
1 3 5 + + +…+
Se porneste de la (2n − ) =4. n -4.n+1 avem:
2 2
1
2 2
1 =4.1 -4.1+1
2 2
3 =4. 2 -4.2+1
2 2
5 =4. 3 -4.3+1
…………………….
(2n − )
2 2
1 =4. n -4n+1
Adunand membru cu membru obtinem:
2 2 2 2
S=4(
1 + 2 +3 +…+ n )-4(1+2+3+…+n)+n=
n( n + 1)(2n + 1) n( n +1) 2n( n +1)(2n + 1)
= 4. -4. +n= -2n(n+1)+n=
6 2 3
3