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Question

Question: Find the nth term for the sequence: \[3,8,15,{\text{ }}24...\]...

Find the nth term for the sequence: 3,8,15, 24...3,8,15,{\text{ }}24...

Explanation

Solution

Check for the difference between the terms that are known, verify if it is an arithmetic progression. Check if they are related by a common factor, to verify if it is a geometric progression. Check for both, to verify if it is a mixture of arithmetic-geometric progression. In case the difference of the terms of the sequence are also a sequence, identify the type and proceed accordingly.

Complete step by step solution:
Let us first subtract the terms from its preceding term and verify if there is a common difference.
83=58 - 3 = 5
158=715 - 8 = 7
2415=924 - 15 = 9
And so on...
Thus, we see that the series doesn’t have a common difference. Rather the differences also form another sequence, that is 5,7,9..5,7,9...
In such cases we cannot apply the formula for the nth term of an arithmetic progression directly.
Let the series in the question have the terms named as T1,T2,T3,T4{T_1},{T_2},{T_3},{T_4}..respectively.
Now, T1{T_1} can be written as,
T1=221=3{T_1} = {2^2} - 1 = 3
Similarly,
T2=321=8 T3=421=15 T4=521=24 . . . Tn=(n+1)21  {T_2} = {3^2} - 1 = 8 \\\ {T_3} = {4^2} - 1 = 15 \\\ {T_4} = {5^2} - 1 = 24 \\\ . \\\ . \\\ . \\\ {T_n} = {(n + 1)^2} - 1 \\\
Thus, Tn=n2+2n+11=n2+2n{T_n} = {n^2} + 2n + 1 - 1 = {n^2} + 2n, where Tn{T_n}is the nth term of the sequence.
Alternate method: Let us consider the sum of the given sequence as, S.
S1=3+8+15+24+..+Tn1+Tn{S_1} = 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}….Consider this equation (i).
Again, S2=0+3+8+15+24+..+Tn1+Tn{S_2} = 0 + 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}……….equation (ii)
Subtracting (ii) from (i), we get
S1S2=0=3+5+7+9+..Tn{S_1} - {S_2} = 0 = 3 + 5 + 7 + 9 + \ldots .. - {T_n}
We see that Tn{T_n} is the only term that is negative on the RHS. Taking it to the LHS, we have
Tn=3+5+7+9...{T_n} = 3 + 5 + 7 + 9...
Now, Tn{T_n} is the sum of odd natural numbers that is equal to n2+2n=n(n+2){n^2} + 2n = n(n + 2)

Note: The “sum of n terms of AP” equals to the sum(addition) of first n terms of the arithmetic sequence. Its adequate n divided by 2 times the sum of twice the primary term, 'a' and therefore the product of the difference between second and first term-'d' also called common difference, and (n-1), where n is the number of terms to be added.