Question

Calcule os seguintes somatórios

Answer 1

a)

É uma soma dos 20 primeiros termos de uma P.A com an = 4n - 1

$a_{1}=4*1-1=4-1=3$
$a_{20}=4*20-1=80-1=79$

$S_{20}=(a_{1}+a_{20})*20/2$
$S_{20}=(3+79)*10$
$S_{20}=82*10$
$S_{20}=820$

b)

$a_{n}=(n+1)/3 \\ a_{1}=(1+1)/3 \\ a_{1}=2/3$

$a_{30}=(30+1)/3=31/3$

$S_{30}=(a_{1}+a_{30})*30/2$
$S_{30}=([2/3]+[31/3])*15$
$S_{30}=([2+31]/3)*15$
$S_{30}=(33/3)*15$
$S_{30}=11*15$
$S_{30}=165$

c)

a₁ + a₂ + a₃ + ... + a₁₉ + a₂₀ + a₂₁ = S₂₁

a₁ = u₂₀ = 2.20 + 3 = 40 + 3 = 43

a₂₁ = u₄₀ = 2.40 + 3 = 80 + 3 = 83

S₂₁ = (a₁ + a₂₁) . 21 / 2
S₂₁ = (43 + 83) . 21 / 2
S₂₁ = 126 . 21 / 2
S₂₁ = 63 . 21
S₂₁ = 1323

Answer 2

Olá, siga a explicação:

A)

$\large {\boxed {\red {\sf \displaystyle \sum_{n=1}^{20} \left ( 4n -1\right ) }}}$

Temos:

$\large {\boxed {\blue {\sf \sum_{}^{}a_n+b_n = \sum_{}^{} a_n + \sum_{}^{} b_n }}}$

Portanto:

$\large {\boxed {\sf \sum_{n=1}^{20 } 4n - \sum_{n=1}^{20}1 }}$

$\large {\boxed { \pink {\sf \sum_{n=1}^{20}4n = 840}}}$

$\large {\boxed {\green {\sf \sum_{n=1}^{20} 1 = 20}}}$

∴ $\large {\boxed {\purple {\sf \bf 820}}}$

B)

$\large {\boxed {\red {\sf \cfrac{n+1}{3} = \cfrac{n}{3} + \cfrac{1}{3} }}}$

$\large {\boxed {\blue {\sf \sum_{}^{}a_n+b_n = \sum_{}^{} a_n + \sum_{}^{} b_n }}}$

$\large {\boxed {\green {\sf \sum _{n=1}^{30}\cfrac{n}{3}+\sum _{n=1}^{30} \cfrac{1}{3}}}}$

$\large {\boxed {\sf \sum_{n=1}^{30} \cfrac{n}{3} = 155 }}$

$\large {\boxed {\purple {\sf \sum_{n=1}^{30} \cfrac{1}{3} = 10 }}}$

∴ $\large {\boxed {\pink {\sf \bf 165}}}$

C)

Regrinha:

$\large {\boxed { \blue {\sf \displaystyle \sum_{k=m}^{n} = \sum_{k=1}^{n} - \sum_{k=1}^{m-1} }}}$

$\large {\boxed { \red {\sf \sum_{n=1}^{40} 2n + 3 - \sum_{n=1}^{19} 2n+3 }}}$

$\large {\boxed {\blue {\sf \sum_{n=1}^{40} 2n+3 = 1760 }}}$

$\large {\boxed {\purple {\sf \sum_{k=1}^{19} 2n+3 = 437 }}}$

∴ $\large {\boxed {\sf \bf 1323 }}$

• Att. MatiasHP