Question

Calculati sumele :
A) 1+2+3...+100
B)1+3+5+...+99
C)2+4+6+...+100
D)5+10+15+...+100

Answer 1

A) 1+2+3+...+n=n(n+1) totul supra 2
asta e formula
si 1+2+3+...+100=100(100+1) totul supra 2=100*101 totl supra 2 
B) 1+3+5+...+99=.... Este de forma 1+3+5+...+2n-1 =n*n; 
99=2n-1 de unde 2n=99+1=100 => n=50 .Deci suma S=50*50=2500; 
C). 2+4+6+...+100=... este de forma 2+4+6+...+2n=n*(n+1); 
=> 2n=100 deci n=5o => S=50*51=2550;
D)5(1+2+3+4+..+20)=5(20*21/2)=5*10*21=50*21=1050

Answer 2

$\displaystyle a).1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} =5050 \\ \\ b).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950-\not2 \times \frac{2450}{\not2} =4950-2450=2500$

$\displaystyle c).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not2 \times \frac{2550}{\not2} =2550 \\ \\ d).5+10+15+...+100=5(1+2+3+...+20)=5 \times \frac{20(20+1)}{2} = \\ \\ =5 \times \frac{20 \times 21}{2} =5 \times \frac{420}{2} =5 \times 210=1050$