Fill in the blanks in the following proof (in attachment), which shows that the sequence defined by the recurrence relation for each integer k ≥ 2
satisfies the following formula. for every integer n ≥ 1
Proof (by mathematical induction):
Suppose , , , ... is a sequence that satisfies the recurrence relation for each integer k ≥ 2, with initial condition .
We need to show that when the sequence , , , ... is defined in this recursive way, all the terms in the sequence also satisfy the explicit formula shown above.
So let the property P(n) be the equation . We will show that P(n) is true for every integer n ≥ 1.

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363