Tricky Series asked in Various Exams

A series in which each term is formed by multiplying the corresponding terms of an A.P. and G.P. is called Arithmetico Geometric series. It is more popularly known as an A.G.P.The general or standard form of such a series is a, (a + d) r, (a + 2 d) r² and so on.

Sum of infinite number of terms of an A.G.P with |r| < 1 is

S ∞ = a / 1– r + rd / (1 – r)²

Where, a = first term
d = common difference of A.P.
r = common ratio of G.P.

Method of Differences:

In some series, the differences of successive terms (T_n and T_n-1) may be in some particular sequence- A.P., G.P., A.G.P. etc.

In these cases, the sum of series can be found by a technique known as "method of differences".

The sum of first n natural numbers-

The sum of squares of first n natural numbers-

The sum of cubes of the first n natural numbers-

Sum of first n odd natural numbers = 1+ 3 + 5 +……+ up to n terms = n².

Sum of first n even natural numbers = 2 + 4 + 6 +……+ up to n terms = n (n + 1).

For any series T_n = S_n – S_n–1

Example 1:

Evaluate 1 + 4/5 + 7/5² + 10/5³ +…… to infinite terms.

Solution:

Here, the series is an A.G.P, where 1, 4, 7, 10…….is an AP, with a = 1 and d = 3and 1, 1/5, 1/5²,….is a GP, with r = 1/5. For an A.G.P.,

S_∞ = a / 1– r + rd / (1 – r)², putting the values we get S_∞ = 35/16.

Example 2:

Find thesum to infinity of the series 1² + 2². x + 3². x² +..., |x| < 1.

Solution:

Though the given series is not an arithmetico-geometric series, however the differences (2² – 1²), (3² – 2²), ……form an AP. So, you can use theMethod of differences.

Let  S = 1 + 4x + 9x² + 16x³ +......∞. Multiply both sides with common ratio x of the GP.
Sx =  x + 4x² + 9x³ +........∞. Now, subtract the two equations
⇒ (1 – x) S = 1 + 3x + 5x² + 7x³ +.....∞ ……..(1)
Now, let R = 1 + 3x + 5x² + 7x³ +......∞, which is an arithmetico-geometric series with a = 1, d = 2 and r = x.
For an A.G.P., sum R = a / 1– r + rd / ( r– r)²
Substituting the values, you get R = 1+ x /(1 – x)²
Substitute R in (1), you get, (1 – x) S = 1+ x /(1 – x)²
⇒ S = 1+ x /(1 – x)³.

Example 3:

Find the sum of the series 2.5 + 5.8 + 8.11 +... upto n terms.

Solution:

This series is formed by multiplying the corresponding terms of sequences 2, 5, 8, ... and 5, 8, 11, ... both of which are A.P.s.

Now, nth term of first A.P. = 2 + (n – 1) 3 = 3n – 1.
and, nth term of second A.P. = 5 + (n – 1) 3 = 3n + 2.
Hence, nth term of the given series, T_n = (3n – 1) (3n + 2) = 9n² + 3n – 2.
Thus, the required sum, 
S_n = T_n = (9 n² + 3n – 2)
= 9  n² + 3 n – 2 = 9 n² + 3 n – 2nn.
⇒ 9 × n (n + 1)(2n + 1)/6 + 3 × n(n + 1)/2 – 2n.
On simplifying ⇒ n (3n² + 6n + 1).

Example 4:

Find the sum of the following series:

Solution:

There are 2007 terms in the given sequence and we want to evaluate S₂₀₀₇. Here, we will apply a shortcut called interpolation.

S₂ = T₁ + T₂ =

On interpolating, we can say S₃ = 4 – 1/4, S₄ = 5 – 1/5 ….so on.
Continuing the same logic S₂₀₀₇ = 2008 – 1/2008.