Please show ho to do b The answer to b is un 3n 2 2n1k I

Please show ho to do (b).........

The answer to (b) is un = 3n - 2 + [2^(n-1)]k I understand that 3n-2 comes from the A.P. ....but I do not understand why we add the G.P [2^(n-1)]k ?

6a. The sums of the terms of a sequence follow the pattern S1 1 k, S. 5 3k, S3 12 7k, S4 22 15 where k E Z. Given that u1 1 k find u2, w3 and u4. Find a general expression for un b.

Solution

S1= 1+ k, S2=5+3k, s3= 12+7k, s4=22+15k ......

u1=1+k

u2=S2-u1=4+2k

u3= S3-(u1 + u2)= 7+4k

u4= S4-(u1+u2+u3)=10+8k

So the sequesnce is

1+k, 4+2k,7+4k,10+8k,.........

As we see that the first term of each term of the sequence is in AP that is

1,4,7,10,........

And nth term of this sequence is = 3n-2

And the second term of the term of the sequence is in GP that is

k,2k,4k,8k,....

And nth term of GP is = a r^(n-1)

here a is the first term which is k here

and r=2

therefore nth term of the GP is = k (2)^(n-1)

Hence u_n= 3n-2 + k(2)^n-1