SATHEE: Binomial Theorem 1 Question 28

Binomial Theorem 1 Question 28

30.

For any odd integer $n \geq 1, n^{3} - (n - 1)^{3} + \dots$ $+ (- 1)^{n - 1} 1^{3} = \dots$

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Answer:

Correct Answer: 30. $(\frac{1}{4} (n + 1)^{2} (2 n - 1))$

Solution:

Since, $n$ is an odd integer, $(- 1)^{n - 1} = 1$

and $n - 1, n - 3, n - 5$ , etc., are even integers, then

$\begin{aligned} n^{3} - & (n - 1)^{3} + (n - 2)^{3} - (n - 3)^{3} + \dots + (- 1)^{n - 1} \cdot 1^{3} \\ = n^{3} + (n - 1)^{3} + (n - 2)^{3} + \dots + 1^{3} \\ - 2 [(n - 1)^{3} + (n - 3)^{3} + \dots + 2^{3})] \\ = Σ n^{3} - 2 \times 2^{3} [\frac{n - 1}{2} + {\frac{n - 3}{2}}^{3} + \dots + 1^{3}] \end{aligned}$

$[∵ n - 1, n - 3, \dots$ , are even integers $]$

$= Σ n^{3} - 16 [Σ (\frac{n - 1}{2})]^{3}$

$= [\frac{n (n + 1)}{2}]^{2} - 16 [\frac{1}{2} (\frac{n - 1}{2}) (\frac{n - 1}{2} + 1)]^{2}$

$= \frac{1}{4} n^{2} (n + 1)^{2} - \frac{16 (n - 1)^{2} (n + 1)^{2}}{4 \times 4 \times 4}$

$= \frac{1}{4} (n + 1)^{2} [n^{2} - (n - 1)^{2}] = \frac{1}{4} (n + 1)^{2} (2 n - 1)$

Binomial Theorem 1 Question 28

30.

Answer:

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Binomial Theorem 1 Question 28

30.

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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