Write the sum without sigma notation. Then evaluate the sum.∑2k=1 40k/k+3Choose the correct answer belowA. (40⋅1/1+3)+(40⋅2/2+3) B. 40k/2+3 C. 40⋅2/2+3 D. (40⋅1.1+3)+(40⋅2/2+3)+(40⋅3/3+3)

Respuesta :

Answer:

A. (40⋅1/1+3)+(40⋅2/2+3)

Step-by-step explanation:

Given that:

[tex]\sum \limits ^2 _{k=1} \dfrac{40 \ k }{k + 3}[/tex]

which can be expressed as:

=[tex]\dfrac{40\times 1 }{1 + 3} + \dfrac{40\times 2 }{2 + 3}[/tex]

=[tex]\dfrac{40 }{4} + \dfrac{80 }{5}[/tex]

= [tex]10+ 16[/tex]

= 26

[tex]\sum \limits ^2 _{k=1} \dfrac{40 \ k }{k + 3} = 26[/tex]