Please help, thank you! Match the sequence and recursive expression to its explicit expression. f(n) = 2n + 10

{2, 4, 6, 8...} ...} f(1) = 2 and f(n) = f(n - 1) + 2 for n >

{12, 14, 16, 18...} f(1) = 12 and f(n) = f(n - 1) + 2 for n > 1

Given:

To find:

The sequence and recursive expression to the given explicit expression.

Solution:

We have,

For n=1,

The value of f(1) is 12.

Similarly,

For n=2,

For n=3,

For n=4,

The required sequence is {12,14,16,18,...}.

The recursive expression of an AP is

where, d is common difference.

Here d=2,

Therefore, the recursive expression is .

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

step-by-step explanation: