Question

Calculati urmatoarele sume :

a) 1+2+3+.......+80
b)2+4+6+........+100
c)1+3+5+.........+99
d)3+7+11+15+........+43

Answer 1

1+2+3+...+80= 80* 81:2=40*81=

Answer 2

$\displaystyle a).1+2+3+...+80= \frac{80(80+1)}{2} = \frac{80 \times 81}{2} = \frac{6480}{2} =3240 \\ \\ b).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not 2 \times \frac{2550}{ \not 2} =2550$

$c).\displaystyle a).1+3+5+...+99 \\ \\ 99=1+(n-1) \times 2 \\ \\ 99=1+2n-2 \\ \\ 2n=99-1+2 \\ \\ 2n=100 \\ \\ n= \frac{100}{2} \\ \\ n=50$

$\displaystyle S_{50}= \frac{2+49 \times 2}{2} \times 50 \\ \\ S_{50}= \frac{2+98}{2} \times 50 \\ \\ S_{50}= \frac{100}{2} \times 50 \\ \\ S_{50}=50 \times 50 \\ \\ S_{50}=2500$

$\displaystyle d).3+7+11+15+...+43 \\ \\ 43=3+(n-1) \times 4 \\ \\ 43=3+4n-4 \\ \\ 4n=43-3+4 \\ \\ 4n=44 \\ \\ n= \frac{44}{4} \\ \\ n=11$

$\displaystyle S_1_1= \frac{6+10 \times 4}{2} \times 11 \\ \\ S_1_1= \frac{6+40}{2} \times 11 \\ \\ S_1_1= \frac{46}{2} \times 11 \\ \\ S_1_1=23 \times 11 \\ \\ S_1_1=253$