# Miscellaneous Progression

Some Important Results:
⋄ 1 + 2 + 3 + … + n = n(n+1)/2  (sum of the first n natural numbers)

⋄ 12 + 22 + 32 + …. + n2= n(n + 1)(2n + 1)/6   (sum of squares of the first n natural numbers)

⋄ 13 + 23 + 33 + …+ n3 = (1 + 2 + 3 + …+ n)2

= [ n(n + 1)/2]2   (sum of cubes of first n natural numbers)

⋄ 1 + x + x2 + x3 +…….. = (1 − x)−1 , if −1 < x < 1

⋄ 1 + 2x + 3x2 +……….. = (1 − x)−2 , if −1 < x < 1

### Method of Differences:

Suppose a1 , a2 , a3 , …….is a sequence such that the sequence a2 − a1 , a3 − a2 , ………is either an A.P. or a G.P. The n th term ‘ an ‘ of this sequence is obtained as follows :

S = a1 + a2 + a3 +……..+ an−1 + an

S = a1 + a2 +………+an-2 + an-1 + an

=> an = a1 + [(a2 − a1) + (a3 − a2) +……+(an − an − 1)]

Since the terms within the brackets are either in an A.P. or in a G.P., we can find the value of an, the nth term. We can now find the sum of the n terms of the sequence as $\large S = \Sigma_{k=1}^{n} a_k$

Illustration Find the sum of series 3 + 8 + 22 + 72 + 266 + 1036 + …….

Solution: 1st difference 5, 14, 50, 194, 770, ….

2nd difference 9 , 36 , 144, 576 , , …..

They are in G.P whose nth term is arn−1 = a4n−1

∴ Tn of the given series will be of the form

Tn = a4n −1 + bn + c

=> T1 = a + b + c = 3

T2 = 4a + 2b + c = 8

T3 = 16a + 3b + c = 22

Solving we have a = 1, b = 2, c = 0.

∴ Tn = 4n − 1 + 2n

∴ Sn = Σ4n − 1 + 2Σn = 1/3(4n − 1) + n(n + 1)

Exercise :

(i) Find the sum of Ist n terms of the series 3 , 6 , 15 , 42 , 123 , …..

(ii) Let Sn denote the sum of first n terms of the sequence 1 , 5 , 14 , 30 , 55 ,…..,

then prove that Sn − Sn−1 = Σn2

(iii) Find the sum of n terms and infinitely many terms of the series

$\large \frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \frac{9}{1^2 + 2^2 + 3^2 + 4^2} + ….$