Miscellanous Series

Now, in this last section, we will look at all the miscellaneous series problems based on Arithmetico-geometric series, Iterative series, User-defined series and others.

Type I: Arithmetico-Geometric Series (AGP)

E.g. 1: Find the sum to infinite terms:

(y + 4y² + 7y³ + 10y⁴ + 13y⁵ + ……………………)

The series given above is an AGP. We can easily identify an AGP as one term of the series will be in AP while the other one will be in GP.

In this series y, y², y³, y⁴, y⁵ are in GP with a common ratio of ‘y’, while the terms 1, 4, 7, 10, 13 and so on are in AP.

So, the standard way of solving AGP problems is to multiply the entire series by the common ratio of the GP and then subtract the newly formed series from the original series. Why do we subtract?

Because on subtracting, we will get a pure geometric progression of infinite terms. And, we know how to find the sum of infinite terms of a GP.

So, let P = y + 4y² + 7y³ + 10y⁴ + 13y⁵ + ……………………[equation (i)]

Now, the common ratio of the GP in AGP is ‘y’. So, multiply the series by ‘y’ .

On multiplying, we get Py = y² + 4y³ + 7y⁴ + 10y⁵ + 13y⁶ +…………………..[equation (ii)]

Now, the trick while subtracting is that leave the first term of the equation (i) as it, start subtracting from the second term onwards as they are like terms.

So, on subtracting, we get:

P – Py = y + (4y² – y²) + (7y³ – 4y³) + (10y⁴ – 7y⁴) +………………….

P(1 – y) = y + 3y² + 3y³ + 3y⁴ + ……………………

Now, starting from the 2^nd term onwards, the series becomes an infinite GP with first term being 3y² and the common ratio being ‘y’.

So, we can find the sum of infinite terms using the formula [a/(1 – r)].

So, on further solving we get,

Type II Iterative Series:

Iterative series means there will be a pattern/cycle/period for the terms. In this section, we will learn how to handle the series in which any term has been represented in terms of its previous terms. So, please learn this concept well, so that the problems can be solved smoothly.

For e.g.: t_n+1 = 2t_n – t_n–1, for n > 1.

In the above example, the previous terms of t_n+1 are t_n and t_{n – 1}, so we can interpret the terms as any new term can be represented as (twice of previous term – yet another previous term).

t_n: previous term of t_n+1.

t_{n – 1}: yet another previous term of t_n+1.

If we can understand this language, we can solve all such types of problems. But, it is not necessary that the always a term will be given in terms of its previous terms. If it is not, then we will have to rearrange it according to our requirements.

E.g. 4: If the sequence t_n+1 + 2t_n–1 = 3t_n, is true for all natural numbers n > 1, then find the value of t₄₀ if t₁ = 1 and t₂ = 2. And also find the sum of those forty terms {t₁ + t₂ + t₃ + t₄ +…………….+ t₄₀}.

Now, this sequence is not arranged in terms of its previous terms, so we will arrange it first.

t_n+1 = 3t_n – 2t_n–1,

Now, we can interpret the sequence as any “new term is difference of thrice of previous term and twice of yet another previous term”.

So, if t₁ = 1

And t₂ = 2

Then t₃ = 3 × t₂ – 2 × t₁ = 6 – 2 = 4

Similarly, t₄ = 3 × t₃ – 2× t₂ = 8

Once, we got the logic, we can write the other terms in a jiffy as t₅ = 16, t₆ = 32,………….and so on.

Again, try to understand one thing, that if the question is asking to find the 40^th term, there must be a pattern or sequence which we can catch.

So, if we try to catch the pattern in the question, it goes like: t₁= 2⁰, t₂ = 2¹, t₃ = 2², t₄ = 2³, t₅ = 2⁴, t₆ = 2⁵. So, t₄₀ should be 2³⁹.

Now, to find the sum of these 40 terms, we can use the formula of sum of the 40 terms of the GP with the first term being 2⁰ and common ratio being 2.

{t₁ + t₂ + t₃ + t₄ +………………+ t₄₀} = {2⁰ + 2¹ + 2² + 2³ + 2⁴ +…………………..+ 2³⁹}.

Alternative method of finding the sum: Whenever there is consecutive series of powers starting from 2⁰, we can use this logic directly and get the answer.

2⁰ + 2¹ = 2² – 1

2⁰ + 2¹ + 2² = 2³ – 1

2⁰ + 2¹ + 2² + 2³ = 2⁴ – 1

2⁰ + 2¹ + 2² + 2³ + 2⁴ = 2⁵ – 1

2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ = 2⁶ – 1

2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ +……………………..+ 2³⁹ = 2⁴⁰ – 1

E.g. 5: If t_n+2 + t_n = t_n+1 for all natural number values of ‘n’ > 0. If t₁ = 3, t₂ = 1, then find out t₁₂₉. And also find out the sum of these terms {t₁ + t₂ + t₃ + t₄ +…………………+ t₁₂₉}.

First of all, we need to see that if the sequence is arranged in the terms of its previous terms.

The previous terms of t_n+2 are t_n+1 and t_n.

So, we can write the sequence as t_n+2 = t_n+1 – t_n.

That means a new term (t_n+2) can be represented in the form of difference of previous term (t_n+1) and yet another previous term (t_n).

So, if t₁ = 3,

t₂ = 1,

t₃ = t₂ – t₁ = 1 – 3 = –2,

t₄ = t₃ – t₂ = –2 – 1 = –3,

t₅ = –3 – (–2) = –1

t₆ = –1 – (–3) = 2

t₇ = 2 – (–1) = 3

So, the first term t₁ is same as t₇, so a cycle of 6 is there in the series. Also, it would have been obvious that if the question was asking to find 129^th term, then cycle/period of terms must be there. Else it will be very cumbersome to find 129^th term.

Now, if a cycle of 6 is being repeated and we want to find t₁₂₉, then we will divide (129 ÷ 6) which gives a remainder of 3.

A remainder of 3 means that t₁₂₉ is the third term of the cycle means equal to t₃.

So, t₁₂₉ = t₃ = –2

Now, to find the sum of the terms {t₁ + t₂ + t₃ + t₄ +………………+ t₁₂₉}, we just need to find the sum of each cycle.

Sum of each cycle is {t₁ + t₂ + t₃ + t₄ + t₅ + t₆} = {3 + 1 – 2 – 3 – 1 + 2} = 0

So, sum of each cycle is 0 and there are 21 cycles in sum of 129 terms, so sum of all 21 cycles will also be zero. But apart from those 21 cycles, three terms are still behind {t₁₂₇, t₁₂₈ and t₁₂₉}.

So, {t₁ + t₂ + t₃ + t₄ +………………+ t₁₂₉} = {t₁₂₇ + t₁₂₈ + t₁₂₉}. And we know t₁₂₇ = t₁, t₁₂₈ = t₂ and t₁₂₉ = t₃.

So, finally sum is {3 + 1 – 2} = 2.

E.g. 10: Find the n^th term of the series whose S_n is given as (2n² + n + 1).

Now, this is the question based on the reverse logic. Up till now, we have solved problems where n^th term has been given and we needed to find out sum of those terms. But, in this case sum is given; we need to find out t_n.

So, now S_n = 2n² + n + 1.

And S_n–1 = 2(n – 1)² + (n – 1) + 1 {just plug-in (n – 1) in place of ‘n’ in S_n}.

On simplifying we get, S_n–1 = 2n² + 2 – 3n.

Finally, t_n = S_n – S_n–1 = 2n² + n + 1 – (2n² + 2 – 3n) = 4n – 1.

If the question was asking to find 10^th term, we could have easily find out since we know the n^th term of the series.

So, t₁₀ = 4 × 10 – 1 = 39.

If the question was not asking to find the n^th term but 10^th term, then we could have directly found out 10^th term by finding out the sum of the 10 terms and then subtracting the sum of the 9 terms.

So, t₁₀ = S₁₀ – S₉.

S_n = 2n² + n + 1, so S₁₀ = 2 × 10² + 10 + 1 = 211.

And S₉ = 2 × 9² + 9 + 1 = 172

And t₁₀ = S₁₀ – S₉ = 211 – 172 = 39.

So, it depends on the question, what is asking. So, accordingly, we can proceed and get the answer. But whatever has been discussed is the crux of all such problems. If we get this, we can solve almost all problems of such type.

E.g. 12: Find the n^th term of the series: 1 × 2 × 4 + 2 × 3 × 5 + 3 × 4 × 6 +…………………..

What we can do to find the n^th term is assume a particular n^th term and check by plug-in values of ‘n’, whether we are getting the terms given in the question.

For e.g., if i say that the n^th term of the above expression is (n) × (n + 1) × (n + 2). You guys will prove it wrong by substituting ‘n’ = 1 in the value of t_n because when we plug-in ‘n’ = 1, we will get t₁ = 1 × 2 × 3, which is not the term given in the question. So, it’s wrong.

Now, you must be confident enough of finding the n^th term. It’s very easy.

So, n^th term (t_n) = (n) × (n + 1) × (n + 3).

If we know the n^th term, we can also find out the sum of these ‘n’ terms easily with the logic learnt so far.

All these problems are based on pattern-catching and familiarity. So do not get overwhelmed by such problems and if all of you solve 5-6 questions of this type, you will get a hang of it and can write the answer in one or two steps.

Let us see one example on the same type again.

8. If the sum to ‘n’ terms of the series is given as 3n² + 4n. Then find the n^th term of the series.

9. There are 8436 steel circular cans, each with a radius of 1 m, stacked in a pile with 1 can on top, 3 cans in the second layer, 6 cans in the third layer, and 10 cans in the fourth layer and so on. Find the number of horizontal layers in the pile.

10. If the n^th term of a series is given as n(n + 3), then find the sum of the first 20 terms of the series.

11. If t_n denotes the n^th term of the series {2, 3, 6, 11, 18,……………….}, then find the 45^th term of the series. Also find the sum of the first 25 term of the series.

12. Find the n^th term of the series:

1 × 2 + 2 × 4 + 3 × 8 + 4 × 16 + 5 × 32 +…………………..

13. Find the sum of the following series:

7 + 26 + 63 + 124 +………………………..+ 728.