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Method of Differences

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

S = a1 + a2 + a3 +…+ an–1 + an                      …… (1)

S = a1 + a2 +…+ an–2 + an–1 + an                    …… (2)

Subtracting (2) from (1), we get, an = a1 + [(a2–a1) + (a3–a2)+…+(an–an–1)].

Since the terms within the brackets are either in an A.P. or 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 S = Σnk=1 ak.

F corresponding to the sequence a1, a2, a3, ……, an, there exists a sequence b0, b1, b2, ……, bn such that ak = bk – bk–1, then sum of n terms of the sequence a1, a2, ……, an is bn – b0.

 

Illustration:
 
        Find the sum of 1st n terms of the series 5, 7, 11, 17, 25, …
 
Solution:
 
        Let    S = 5 + 7 + 11 + 17 + 25 +…+ tn.
        Also   S = 5 + 7 + 11 + 17 +… + tn–1 + tn.
        Subtracting we get
        0 = 5 + 2 + 4 + 6 + 8 +…+ nth term – tn
 
        ⇒ tn = 5 + 2 {n(n–1)/2}.
        or tn = n2 – n + 5
         Sn = Σn2 – Σn + 5Σ 1 = n(n+1)(2n+1)/6 – n(n+1)/2 + 5n
                = n/6 {(n + 1)(2n + 1) –3(n + 1) + 30}
                = n/6 (2n2 + 28).
 
Illustrations based on Vn method
 
Illustration:
 
        Find the sum of the series 1/1.2.3.4 + 1/2.3.4.5 +...... n terms.
 
Solution:
 
Let Tr =1/r(r+1)(r+2)(r+3) = 1/3 – [(r+3)–r]/r(r+1)(r+2)(r+3)
        =1/3 [1/r(r+1)(r+2) – 1/(r+1)(r+2)(r+3)] = 1/3 [Vr–1 – Vr]
 Sn = Σnr=1 Tr = 1/3Σnr=1 [Vr–1 – Vr] = 1/3 [V0 – Vn] = 1/3 [1/1.2.3 – 1/(n+1)(n+2)(n+3)]
 
Illustration:
 
        Find the sum of the series 1.2.3.4.5 + 2.3.4.5.6 + …… n terms.
 
Solution:
 
     Let Tr = r(r + 1)(r + 2)(r + 3)(r + 4)
        = 1/6 [(r+5)–(r–1)] r(r + 1)(r + 2)(r + 3)(r+ 4)
        = 1/6 [r(r+1)…(r+5)–(r+4)] =  [Vr – Vr–1]

   S = 1/6 Σnr=f [Vr – Vr–1] = 1/6 [Vn – V0]

        = 1/6[n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)].
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