Miscellaneous progressions:

Some most Important Results:

• 1 + 2 + 3+...+n =    (summation of the first n natural numbers)
• 1² + 2² + 3² + ... + n² =   (summation of squares of the first n natural numbers)
• 1³+2³+3³+...+n³=  =(1+2+3+...+n)² (summation 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

Consider a₁, a₂, a₃, .......is a linear sequence such that the sequence a₂ - a₁, a₃ - a₂, .........is either an A.P. or a G.P. The n^th term 'a_n' of this sequence is calculated as follows :

S = a₁ + a₂ + a₃ +........+a_n-1 + a_n

S = a₁ + a₂ +.........+a_n-2 + a_n-1 + a_n

=>  a_n = a₁ + [(a₂ - a₁) + (a₃ - a₂) +......+(a_n - a_{n - 1})]

Since the terms within the brackets are either in an A.P. or in a G.P., we may calculate the value of a_n, the n^th term. We may now calculate the sum of the n terms of the linear sequence as 

Problem 14:       Calculate the sum of n terms of the series 1+ 5 + 11 + 19 + 29+ ..........

Solution:             Here the difference of successive terms of the series is 4, 6, 8, 10... which are in AP.

                              Consider S_n = 1 + 5 + 11 + 19 + ......+ t_{n -1} + t_n                                  ...(1)

                              And S_n =      1 + 5 + 11 + 19 + ......+ t_{n -1} + t_n                          ...(2)

                              Subtracting (2) from (1), we obtain

                              0 = 1 + {4 + 6 + 8 + .....to (n -1) terms} -t_n  

                              or t_n  = 1 + {4 + 6 + 8 + .....to (n -1) terms}

                              = 1 + n-1/2 {2.4 + (n -1 -1)2} = 1 + (n -1) (n + 2) = 1 + n² +n -2 = n² + n - 1

                             ∴ S_n = ∑ t_n =∑ (n² + n -1) =∑ n² +∑ n - ∑ 1

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