Question

PLEASE HELP ME!

Answer 1

Step-by-step explanation:

$1.\sum_{i=1}^{5}3i$

The simplest method is "brute force".  Calculate each term and add them up.

∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)

∑ = 3 + 6 + 9 + 12 + 15

∑ = 45

$2.\sum_{k=1}^{4}(2k)^{2}$

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

∑ = 4 + 16 + 36 + 64

∑ = 120

$3.\sum_{k=3}^{6}(2k-10)$

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

∑ = -4 + -2 + 0 + 2

∑ = -4

4. 1 + 1/4 + 1/16 + 1/64 + 1/256

This is a geometric sequence where the first term is 1 and the common ratio is 1/4.  The nth term is:

a = 1 (1/4)ⁿ⁻¹

So the series is:

$\sum_{j=1}^{7}(\frac{1}{4})^{j-1}$

5. -5 + -1 + 3 + 7 + 11

This is an arithmetic sequence where the first term is -5 and the common difference is 4.  The nth term is:

a = -5 + 4(n−1)

a = -5 + 4n − 4

a = 4n − 9

So the series is:

$\sum_{j=1}^{5}(4j-9)$