PLEASE HELP ME!

1. What is the sum of the series?

∑i=153i

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2. What is the sum of the series?

∑k=14(2k)2

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3. What is the sum of the series?

∑k=36(2k−10)

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4. Which answer represents the series in sigma notation?

1+14+116+164+1256+11024+14096

∑j=174j−1

∑j=1714j+1

∑j=17(14)j−1

∑j=1714(j+1)

5. Which answer represents the series in sigma notation?

−5+(−1)+3+7+11

∑j=155j−1

∑j=15−5⋅(15)j−1

∑j=15(4j−9)

∑j=15(−6j+1)

The sigma notation allows writing an expression in a compact form using the upper and lower boundaries of the variable, then the actual expression. The solutions to the problems are as follows.

1.)

; values of i ranges from 1 - 5 ;

2.)

; values of k ranges from 1 - 5 ;

= (2×1)² + (2×2)² + (2×3)² + (2×4)²

= 4 + 16 + 36 + 64

= 120

3.)

; values of k ranges from 3 - 6;

= (2(3) - 10) + (2(4) - 10) + (2(5) - 10) + (2(6) - 10)

= - 4 - 2 + 0 + 2

= - 4

4.)

The common ratio, r = r2 ÷ r1 = r3 ÷ r2 = r4 ÷ r3 =...

r = 1/4 ÷ 1/1 = 1/4

The nth term of a geometric sequence :

Hence, the series in sigma notation can be expressed thus :

5.)

This is an arithmetic series :

d = -1 - (-5) = -1 + 5 = 4

nth term = a + (n-1)d ; n = number of terms ; a = first term

General expression :

-5 + (n - 1)4

-5 + 4n - 4

4n - 9

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Step-by-step explanation:

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

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

∑ = 4 + 16 + 36 + 64

∑ = 120

∑ = (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:

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:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.