JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Sequences and Series

Let

S_{n} = \sum_{k = 1}^{4 n} (- 1)^{\frac{k ( k + 1 )}{2}} k^{2}

. Then

S_{n}

can take value(s)

(A)1056✓ Correct
(B)1088
(C)1120
(D)1332✓ Correct

Answer: (A), (D)

Solution

The sum can be regrouped as

S_{n} = - 1^{2} - 2^{2} + 3^{2} + 4^{2} - 5^{2} - 6^{2} + \dots + (4 n)^{2}

. Pairing terms:

S_{n} = (3^{2} - 1^{2}) + (4^{2} - 2^{2}) + (7^{2} - 5^{2}) + (8^{2} - 6^{2}) + \dots

for

2 n

pairs each. Each pair of the form

(4 k - 1)^{2} - (4 k - 3)^{2} = 16 k - 8

and

(4 k)^{2} - (4 k - 2)^{2} = 16 k - 4

. Summing these patterns gives

S_{n} = 4 n (4 n + 1)

. For

n = 8

S_{8} = 4 (8) (33) = 1056

. For

n = 9

S_{9} = 4 (9) (37) = 1332

. Thus options A and D are correct.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · One or more correct options

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