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 n^th term, of this sequence is obtained as follows:

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

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

Subtracting (2) from (1), we get, 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 a G.P., we can find the value of an, the n^th term. We can now find the sum of the n terms of the sequence as S = Σⁿ_k=1 a_k.

F corresponding to the sequence a₁, a₂, a₃, ……, a_n, there exists a sequence b₀, b₁, b₂, ……, b_n such that a_k = b_k – b_k–1, then sum of n terms of the sequence a₁, a₂, ……, a_n is b_n – b₀.

Illustration:

Find the sum of 1^st n terms of the series 5, 7, 11, 17, 25, …

Solution:

Let S = 5 + 7 + 11 + 17 + 25 +…+ t_n.

Also S = 5 + 7 + 11 + 17 +… + t_n–1 + t_n.

Subtracting we get

0 = 5 + 2 + 4 + 6 + 8 +…+ n^th term – t_n

⇒ t_n = 5 + 2 {n(n–1)/2}.

or t_n = n² – n + 5

⇒ S_n = 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 T_r =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 [V_r–1 – V_r]

⇒ S_n = Σⁿ_r=1 T_r = 1/3Σⁿ_r=1 [V_r–1 – V_r] = 1/3 [V₀ – V_n] = 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 T_r = 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)] = [V_r – V_r–1]

⇒ S = 1/6 Σⁿ_r=f [V_r – V_r–1] = 1/6 [V_n – V₀]

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

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