Question

a) 2=4=6+.....+100
b) 3+7+11+15+.......+43

Answer 1

 

$\displaystyle\bf\\a)~~\text{Consider ca ai vrut sa scrii: }~2+4+6+...+100\\\\2+4+6+...+100=?\\\\\text{Calculam numarul de termeni:}\\\\n=\frac{100-2}{2}+1=\frac{98}{2}+1=49+1=50~de~termeni\\\\\text{Calculam suma:}\\\\2+4+6+...+100=\\\\=\frac{n(ultimul~termen + primul~termen)}{2}=\\\\=\frac{50(100 + 2)}{2}=\frac{50\times102}{2}=50\times51=\boxed{\bf2550}$

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$\displaystyle\bf\\b)\\\\3+7+11+15+.......+43\\\\\text{Calculam numarul de termeni:}\\\\n=\frac{43-3}{4}+1=\frac{40}{4}+1=10+1=11~termeni\\\\\text{Calculam suma:}\\\\3+7+11+15+...+43=\\\\=\frac{n(ultimul~termen + primul~termen)}{2}=\\\\=\frac{11(43 + 3)}{2}=\frac{11\times46}{2}=11\times23=\boxed{\bf253}$

 

Answer 2

Răspuns:

2(1+2+3+,,,+50)=2 (1+50)×50/2=51×50=2550 ;   4x-1 =43 4x=44 ; x=44/4=11 termeni ;  (3+43)×11/2=23×11=253

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