Question

calculati a la puterea 2 si a la puterea 3,unde a=(1+0,1)+(1+0,2)+(1+0,3)+...+(1+0,9)

Answer 1

[tex]\it (1+0,1)+(1+0,2)+(1+0,3)+ ... +(1+0,9)= \\\;\\ = \it\underbrace{ \it 1+1+1 + ... +1}_{9\ termeni} + \it \dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+ ... +\dfrac{9}{10} = \\\;\\ = \it9 + \dfrac{\it1+2+3 + ...+9}{\it 10} = \it9+ 4,5 =13,5[/tex]