SATHEE: Binomial Theorem 1 Question 30

Binomial Theorem 1 Question 30

32.

Prove that $\sum_{r = 1}^{k} (- 3)^{r - 1}^{3}^{n} C_{2 r - 1} = 0$ , where $k = (3 n) / 2$ and $n$ is an even positive integer.

(1993, 5M)

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Solution:

Since, $n$ is an even positive integer, we can write

$n = 2 m, m = 1, 2, 3, \dots$

Also, $k = \frac{3 n}{2} = \frac{3 (2 m)}{2} = 3 m ∴ S = \sum_{r = 1}^{3 m} (- 3)^{r - 1} \cdot^{6 m} C_{2 r - 1}$

i.e. $S = (- 3)^{0}^{6 m} C_{1} + (- 3)^{6 m} C_{3} + \dots$ $+ (- 3)^{3 m - 1} \cdot^{6 m} C_{3 m - 1} \dots \dots . (i)$

From the binomial expansion, we write

$\begin{aligned} (1 + x)^{6 m} & =^{6 m} C_{0} +^{6 m} C_{1} x +^{6 m} C_{2} x^{2} + \dots \\ ^{6 m} C_{6 m - 1} x^{6 m - 1} +^{6 m} C_{6 m} x^{6 m} \dots \dots . (i i) \\ (1 - x)^{6 m} = & ^{6 m} C_{0} +^{6 m} C_{1} (- x) +^{6 m} C_{2} (- x)^{2} + \dots \\ +^{6 m} C_{6 m - 1} (- x)^{6 m - 1} +^{6 m} C_{6 m} (- x)^{6 m} \dots \dots . (i i i) \end{aligned}$

On subtracting Eq. (iii) from Eq. (ii), we get

$(1 + x)^{6 m} - (1 - x)^{6 m} = 2 [^{6 m} C_{1} x +^{6 m} C_{3} x^{3}$ $+^{6 m} C_{5} x^{5} + \dots +^{6 m} C_{6 m - 1} x^{6 m - 1}]$

$\Rightarrow \frac{(1 + x)^{6 m} - (1 - x)^{6 m}}{2 x} =^{6 m} C_{1} +^{6 m} C_{3} x^{2} +^{6 m} C_{5} x^{4} + \dots$ $+^{6 m} C_{6 m - 1} x^{6 m - 2}$

$Let x^{2} = y$

$\Rightarrow \frac{(1 + \sqrt{y})^{6 m} - (1 - \sqrt{y})^{6 m}}{2 \sqrt{y}} =^{6 m} C_{1} +^{6 m} C_{3} y$

$+^{6 m} C_{5} y^{2} + \dots +^{6 m} C_{6 m - 1} y^{3 m - 1}$

For the required sum we have to put $y = - 3$ in RHS.

$\begin{aligned} ∴ S & = \frac{(1 + \sqrt{- 3})^{6 m} - (1 - \sqrt{- 3})^{6 m}}{2 \sqrt{- 3}} \\ = \frac{(1 + i \sqrt{3})^{6 m} - (1 - i \sqrt{3})^{6 m}}{2 i \sqrt{3}} \dots \dots . (i v) \end{aligned}$

Let $z = 1 + i \sqrt{3} = r (\cos θ + i \sin θ)$

$\Rightarrow r = | z | = \sqrt{1 + 3} = 2$

and $θ = π / 3$

Now, $z^{6 m} = [r (\cos θ + i \sin θ)]^{6 m}$

$= r^{6 m} (\cos 6 m θ + i \sin 6 m θ)$

Again, $\bar{z} = r (\cos θ - i \sin θ)$

and $(\bar{z})^{6 m} = r^{6 m} (\cos 6 m θ - i \sin 6 m θ)$

$\Rightarrow z^{6 m} - {\bar{z}}^{6 m} = r^{6 m} (2 i \sin 6 m θ) \dots \dots . (v)$

From Eq. (i),

$\begin{aligned} S & = \frac{z^{6 m} - {\bar{z}}^{6 m}}{2 i \sqrt{3}} = \frac{r^{6 m} (2 i \sin 6 m θ)}{2 i \sqrt{3}} \\ = \frac{2^{6 m} \sin 6 m θ}{\sqrt{3}} \\ = 0 as m \in z, and θ = π / 3 \end{aligned}$

Binomial Theorem 1 Question 30

32.

Solution:

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Binomial Theorem 1 Question 30

32.

Solution:

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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