Question

Evaluate the following:

(a) Σ(2i-1) i=1 to 12
(b) Σ80 i=1 to 2
Expand the following:
(a) Σ(2xi 3yi) i=1 to 3

Express the following in the Σ notation:
(a) x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6
(b) (x12 y12) (x22 y22) ______ (xk2 yk2)

Answer 1

For question a, the sum Σ(2i-1) from i=1 to 12 evaluates to 169. For question b, the sum Σ80 from i=1 to 2 evaluates to 160. For question c, the expansion of Σ(2xi 3yi) from i=1 to 3 involves multiplying each term by the corresponding combination of xi and yi. For question d, the expression x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 can be written in Σ notation as Σ(i * xi) from i=1 to 6. For question e, the expression (x12 y12) (x22 y22) .... (xk2 yk2) can be expressed in Σ notation as Π((xi)^(2 * i) * (yi)^(2 * i)) from i=1 to k.

Step-by-step explanation:

Question a:

To evaluate Σ(2i-1) from i=1 to 12, we sum up the expression 2i-1 for each value of i from 1 to 12. Here are the steps:

1. Plug in the first value of i (i.e., 1) into the expression 2i-1: 2(1)-1 = 1. This is the first term.
2. Continuing in the same way, we find the next terms by plugging in 2, 3, and so on, up to 12 into the expression.
3. Finally, add up all the terms: 1 + 3 + 5 + ... + 23 = 169. This is the evaluation of the sum.

Question b:

To evaluate Σ80 from i=1 to 2, we sum up the constant value 80 for each value of i from 1 to 2. Here are the steps:

1. Plug in the first value of i (i.e., 1) into the expression 80: 80. This is the first term.
2. Continuing in the same way, we find the next term by plugging in 2 into the expression, which is also 80.
3. Finally, add up both terms: 80 + 80 = 160. This is the evaluation of the sum.

Question c:

To expand Σ(2xi 3yi) from i=1 to 3, we multiply each term in the series by the corresponding combination of xi and yi. Here is the expansion:

1. Using the first value of i (i.e., 1), we have 2(1)x1 3(1)y1 = 2x1 + 3y1.
2. Continuing in the same way, we find the expansions for all other terms.

Question d:

To express the given expression x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 in Σ notation, we can write it as Σ(i * xi) from i=1 to 6. Here is the expression in Σ notation:

x1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 = Σ(i * xi) from i=1 to 6.

Question e:

The expression (x12 y12) (x22 y22) .... (xk2 yk2) can be expressed in Σ notation as Π((xi)^(2 * i) * (yi)^(2 * i)) from i=1 to k. Here is the expression in Σ notation:

(x12 y12) (x22 y22) .... (xk2 yk2) = Π((xi)^(2 * i) * (yi)^(2 * i)) from i=1 to k.