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)

asked Aug 15, 2021 126k views

3 votes

by Es

6.5k points

Please log in or register to add a comment.

4 votes

Answer:

A. (40⋅1/1+3)+(40⋅2/2+3)

Explanation:

Given that:

$\sum \limits ^2 _(k=1) (40 \ k )/(k + 3)$

which can be expressed as:

=
(40* 1 )/(1 + 3) + (40* 2 )/(2 + 3)

=
(40 )/(4) + (80 )/(5)

=
10+ 16

= 26

∴

$\sum \limits ^2 _(k=1) (40 \ k )/(k + 3) = 26$

Tracee answered Aug 21, 2021

6.3k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

6.5m questions

8.7m answers