SATHEE: Sequences and Series 2 Question 10

Sequences and Series 2 Question 10

11.

If $S_{n} = \sum_{2}^{4 n} (- 1)^{\frac{k (k + 1)}{2}} k^{2}$ . Then, $S_{n}$ can take value(s)

(a) 1056

(b) 1088

(d) 1332

(2013 Adv.)

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Answer:

Correct Answer: 11. (a,d)

Solution:

PLAN: Convert it into differences and use sum of nterms of an AP,

$i.e. S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

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

$\begin{aligned} = - (1)^{2} - 2^{2} + 3^{2} + 4^{2} - 5^{2} - 6^{2} + 7^{2} + 8^{2} + \dots \\ = (3^{2} - 1^{2}) + (4^{2} - 2^{2}) + (7^{2} - 5^{2}) + (8^{2} - 6^{2}) + \dots \\ = \underset{n terms}{\underset{⏟}{2 {(4 + 6 + 12 + \dots)}_{n terms} + \underset{⏟}{(6 + 14 + 22 + \dots)}}} \\ = 2 \frac{n}{2} 2 \times 4 + (n - 1) 8 + \frac{n}{2} 2 \times 6 + (n - 1) 8 \\ = 2 [n (4 + 4 n - 4) + n (6 + 4 n - 4)] \\ = 2 [4 n^{2} + 4 n^{2} + 2 n] = 4 n (4 n + 1) \end{aligned}$

Here, $1056 = 32 \times 33, 1088 = 32 \times 34$ ,

$1120 = 32 \times 35, 1332 = 36 \times 37$

1056 and 1332 are possible answers.

Sequences and Series 2 Question 10

11.

Answer:

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Sequences and Series 2 Question 10

11.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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