Find the 7th term of the sequence of 10, 23, 60, 169, 494,.... Please help.?

Let the n-th term of this sequence be a(n) , I found that

a(1) = 6*3^(1-1) + 1 + 3 = 10

a(2) = 6*3^(2-1) + 2 + 3 = 23

a(3) = 6*3^(3-1) + 3 + 3 = 60

a(4) = 6*3^(4-1) + 4 + 3 = 169

a(5) = 6*3^(5-1) + 5 + 3 = 494

So , I think a(n) = 6*3^(n-1) + n + 3 .

Therefore , a(7) = 6*3^(7-1) + 7 + 3 = 4384 .

The only thing I'm coming up with is:

23 = 3 x 10 - 7 (term # 2)

60 = 3 x 23 - 9 (term # 3)

169 = 3 x 60 - 11 (term # 4)

494 = 3 x 169 - 13 (term # 5)

Looks like you have to keep this pattern

going until you hit the 10th term

23 = 3 x 10 - 7 (term # 2)

60 = 3 x 23 - 9 (term # 3)

169 = 3 x 60 - 11 (term # 4)

494 = 3 x 169 - 13 (term # 5)

follow this pattern until you reach the 7th term

Well it's not an arithmetic sequence (the differences are growing) and it's not a geometric sequence (the ratio between terms is not a constant).

Let's take a look at the differences and see if there's a pattern.

13, 37, 109, 325,...

Hmm. Each difference is a little less than 3 times the previous one.

13 * 3 = 39. Subtract 2 to get 37

37 * 3 = 111. Subtract 2 to get 109

109 * 3 = 327. Subtract 2 to get 325.

So the next difference will be 327*3 - 2, and the next difference will be 3 times that minus 2.

Let

P(n): 10.....23.....60.....169.....494......

T(n):......13.....37.....109.....325....

u(n)............24.....72.......216...

=>

P(5)=494

T(4)=325

u(3)=216=24*3^2=u(1)3^2

=>

p(5)=p(4)+T(3)+u(1)3^2

=>

p(n)=p(n-1)+T(n-2)+u(1)3^(n-3)

is the general expression

=>

p(6)=494+325+24(3^3)=1467

=>

p(7)=1467+(1467-494)+24(3^4)=4384

2848, because if a(n) = 4n^4 - 32n^3 + 104n^2 - 135n + 69, then a(1) = 10, a(2) = 23, a(3) = 60, a(4) = 169, a(5) = 494, .... , so just plug in n=7.

10, 23,60,169,494,1477,4404

first order differences between terms are 13,37,109,335,983,2927

Second Order difference between terms 24,72,216,648,1944

second order differences are multiples of three so 216*3 = 648 and 648*3= 1944

335+648+494 = 1477 this 6th term

1944+983+1477=4404

f(n) = 10,23, 60,169, 494,1467, 4384

d1(n) = 13,37,109,325, 973,2917, 8749

d2(n) = 24,72,216,648,1944,5832,17496.

The breakthrough occurs when one recognizes

that d2(n) = 24[3^(n-1)], n=1,2,3,..... Then one

reverse generates d1(n) and f(n).

I hate the OP

10,(10.3-7),(23.3-9), (60.3-11),169.3-13),(494.3-15),(1494,3-17),...

10,23,60,169,494,1494,5390