Question

23*** Calculaţi: a1+2+3+...+ 200; 2+4+6+ ... + 200; b1+3+5+...+ 101; d) 1+4+7+ ... + 301.​

Answer 1

Răspuns:

Explicație pas cu pas:

a)   1+2+3+.......+200 = 200×(1+200) : 2 = 200×201:2 = 20100

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2+4+6+.......+200 = 10 100

=2×1+2×2+2×3+2×4+.....+2×100 =

= 2×(1+2+3+4+......+100) =

= 2×100×(1+100):2 =

= 100×101 =

= 10 100

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b)   1+3+5+.......+101 =

(101-1) : 2 + 1 = 51 termeni impari consecutivi are suma

= 51 × (1+101) : 2 =

= 51 × 102 : 2 =

= 51 × 51 =

= 2 601

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d)  1+4+7+......+301

4-1=3,  7-4=3 ⇔ r = 3

(301 -1 ) : 3 + 1 = 101 termeni are suma

= 101 × ( 1+301) : 2 =

= 101 × 302 : 2 =

= 30 502 : 2 =

= 15 251  

Answer 2

$\it a)\ \ \ 1+2+3+\ ...\ +200=\dfrac{200\cdot201}{2}=100\cdot201=20100\\ \\ \\ b)\ \ \ 1+3+5+\ ...\ +101=51^2=2601\\ \\ \\ d)\ \ a_n=a_1+(n-1)r \Rightarrow 301=1+(n-1)\cdot3|_{-1}\Rightarrow 300=(n-1)\cdot3|_{:3}\Rightarrow\\ \\ \\ \Rightarrow100=n-1 \Rightarrow n=101\\ \\ 1+4+7+\ ...\ +301=\dfrac{(1+301)\cdot101}{2}=\dfrac{302\cdot101}{2}=151\cdot101=$