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 + ……