A lvl H2 Maths: Sequences

Question from http://www.sgforums.com/forums/2297/topics/348959

If S_n = 1 - 4 + 9 - 16 + ... + (n)² (-1)⁽ⁿ⁺¹⁾ and given that S_2r+1 - S_2r-1 = 4r + 1, show that S_2n+1 = (n+1)(2n+1).

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Answer:

S₁ = 1
Using the method of differences,
S_2n+1 - S₁
= S_2n+1 - S_2n-1
+ S_2n-1 - S_2n-3
+ S_2n-3 - S_2n-5
+ ..... - .....
+ S₅ - S₃
+ S₃ - S₁

= A.P. of n terms with first term 5 and common difference 4 (T₁=5, T₂=9)
= (n/2) (2(5) + (n-1)(4))
= n (5 + 2n - 2)
= 2n² + 3n

Thus,
S_2n+1 - S₁ = 2n² + 3n
S_2n+1 - 1 = 2n² + 3n
S_2n+1 = 2n² + 3n + 1
S_2n+1 = (n + 1)(2n + 1) (shown)

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