Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative approach such as that used in Example 10. a) an = −an−1, a0 = 5 b) an = an−1 + 3, a0 = 1 c) an = an−1 − n, a0 = 4 d) an = 2an−1 − 3, a0 = −1 e) an = (n + 1)an−1, a0 = 2 f ) an = 2nan−1, a0 = 3 g) an = −an−1 + n − 1, a0 = 7

< 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)