Method of Difference – Sequences and Series

Here you will learn method of difference in sequences and series with examples.

Let’s begin –

Method of Difference

Some times the \(n^{th}\) term of a sequence or a series can not be determined by the method, we have discussed earlier. So we compute the difference between the successive terms of given sequence for obtained the \(n^{th}\) terms.

If \(T_1\), \(T_2\), \(T_3\),…….,\(T_n\) are the terms of a sequence then some times the terms \(T_2\) – \(T_1\),
\(T_3\) – \(T_1\)…… constitute an AP/GP. \(n^{th}\) term of the series is determined & the sum to n terms of the sequence can easily be obtained.

Case 1 :

(a) If difference series are in A.P., then

Let \(T_n\) = \(an^2\) + bn + c, where a, b, c are constant

(b) If difference of difference series are in A.P.

Let \(T_n\) = \(an^3\) + \(bn^2\) + cn + d, where a, b, c, d are constant

Case 2 :

(a) If difference are in G.P., then

Let \(T_n\) = \(ar^n\) + b, where r is common ratio & a, b are constant

(b) If difference of difference are in G.P., then

Let \(T_n\) = \(ar^n\) + bn + c, where r is common ratio & a, b, c are constant

Determine constant by putting n = 1, 2, 3 ……. n and putting the value of \(T_1\), \(T_2\), \(T_3\)……. and sum of series \(S_n\) = \({\sum}T_n\)

Example : Find the sum to n terms of the series : 3 + 15 + 35 + 63 + …..

Solution : The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. Clearly,these differences are in AP.

Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series

Then, \(S_n\) = 3 + 15 + 35 + 63 + ….. + \(T_{n-1}\) + \(T_n\) ……(i)

Also, \(S_n\) =     3 + 15 + 35 + ….. + \(T_{n-1}\) + \(T_n\) ……(ii)

Subtracting (ii) from (i), we get

0 = 3 + [12 + 20 + 28 + …… + \(T_{n-1}\) + \(T_n\)] – \(T_n\)

\(\implies\) \(T_n\) = 3 + \((n-1)\over 2\){2*12+(n-1-1)*8} = 3 + (n-1){12+4n-8}

\(\implies\) \(T_n\) = 3 + (n-1)(4n+4) = \(4n^2 – 1\)

\(\therefore\) \(S_n\) = \(\sum_{k=1}^{n}\) \(T_k\) = \(\sum_{k=1}^{n}\) (\(4k^2 – 1\)) = 4\(\sum_{k=1}^{n}\)\(k^2\) – \(\sum_{k=1}^{n}\) 1

\(\implies\) \(S_n\) = 4{\(n(n+1)(2n+1)\over 6\)} – n = \(n\over 3\) (\(4n^2 + 6n – 1\))


Related Questions

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + …….

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……

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