Precalculus, Chapter 9, 9.4, Section 9.4, Problem 42

3,-9/2,27/4,-81/8,....
The terms of the above sequence can be written as,
3,3(-3/2),3(-3/2)(-3/2),3(-3/2)(-3/2)(-3/2),.....
So from the above we can see that the each consecutive term of the sequence is multiplied by a common ratio (-3/2).
So the nth term a_n of the sequence can be written as 3*(-3/2)^(n-1)
In order to find the formula for the first n terms of the sequence, Let's first write down the first few sums of the sequence.
S_1=3
S_2=3+(-9/2)=-3/2=(3[1-(-3/2)^2])/(1-(-3/2))=(3[1-(-3/2)^2])/(1+3/2)=6/5[1-(-3/2)^2]
S_3=3+(-9/2)+27/4=21/4=(3[1-(-3/2)^3])/(1-(-3/2))=(3[1-(-3/2)^3])/(1+3/2)=6/5[1-(-3/2)^3]
S_4=3+(-9/2)+27/4+(-81/8)=-39/8=(3[1-(-3/2)^4])/(1-(-3/2))= (3[1-(-3/2)^4])/(1+3/2)=6/5[1-(-3/2)^4]
From the above , it appears that the formula for the sum of the k terms of the sequence is,
S_k=6/5[1-(-3/2)^k]
Let's check the above formula for n=1,
S_1=6/5[1-(-3/2)^1]=6/5(1+3/2)=6/5(5/2)=3
So the formula is verified for n=1.
Now let's assume that the formula is valid for n=k and we have to show that it is valid for n=k+1.
S_(k+1)=3+(-9/2)+27/4+(-81/8)+......+3(-3/2)^(k-1)+3(-3/2)^(k+1-1)
S_(k+1)=S_k+3(-3/2)^k
S_(k+1)=6/5[1-(-3/2)^k]+3(-3/2)^k
S_(k+1)=6/5-6/5(-3/2)^k+3(-3/2)^k
S_(k+1)=6/5+(-3/2)^k(3-6/5)
S_(k+1)=6/5+(-3/2)^k((15-6)/5)
S_(k+1)=6/5+9/5(-3/2)^k
S_(k+1)=6/5[1+9/6(-3/2)^k]
S_(k+1)=6/5[1+3/2(-3/2)^k]
S_(k+1)=6/5[1-(-3/2)(-3/2)^k]
S_(k+1)=6/5[1-(-3/2)^(k+1)]
So, the formula is true for n=k+1 also.
Hence the formula for the sum of the n terms of the sequence is,
S_n=6/5[1-(-3/2)^n]