✿————✦————✿————✦————✿
The answer is: <u>2(k2−4k)(2c+5)</u>
✿————✦————✿————✦————✿
Step:
* Consider 2ck2+5k2−8ck−20k. Do the grouping 2ck2+5k2−8ck−20k=(2ck2+5k2) +(−8ck−20k), and factor out k2 in the first and −4k in the second group.
* Factor out the common term 2c+5 by using the distributive property.
* Rewrite the complete factored expression.
✿————✦————✿————✦————✿
Answer: a) add n+1 to the previous term
b) add the previous two terms
d) subtract n+1 from the previous term
e) multiply the previous term by 3
f) subtract 2 from previous term then add 5 to the next term
<u>Step-by-step explanation:</u>
a) 1, 3, 6, 10
∨ ∨ ∨
+2 +3 +4 The next term is 10 +5 = 15
b) 1, 2, 3, 5
∨ ∨ ∨
=3 =5 =8 The next term is 5 + 8 = 13
d) 8, 7, 5, 2
∨ ∨ ∨
-1 -2 -3 zThe next term is 2 - 4 = -2
e) 1, 3, 9, 27
∨ ∨ ∨
×3 ×3 ×3 The next term is 27 × 3 = 81
f) 49, 47, 52, 50, 55
∨ ∨ ∨ ∨
-2 +5 -2 +5 The next term is 55 - 2 = 53
The following term is 53 + 5 = 58
