SATHEE: Binomial Theorem 2 Question 13

Binomial Theorem 2 Question 13

16.

If $n$ is a positive integer and

${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + \dots + a_{2 n} x^{2 n}$

Then, show that, $a_{0}^{2} - a_{1}^{2} + \dots + a_{2 n}^{2} = a_{n}$ .

(1994, 5M)

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Solution:

${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + \dots + a_{2 n} x^{2 n}$ …….(i)

Replacing $x$ by $- 1 / x$ , we get

$1 - \frac{1}{x} + {\frac{1}{x^{2}}}^{n} = a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} - \frac{a_{3}}{x^{3}} + \dots + \frac{a_{2 n}}{x^{2 n}}$ …….(ii)

Now, $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + \dots + a_{2 n}^{2} =$ coefficient of the term independent of $x$ in

$\begin{aligned} [a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}] \\ \times [a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} - \dots + \frac{a_{2 n}}{x^{2 n}}] \end{aligned}$

$=$ Coefficient of the term independent of $x$ in

${(1 + x + x^{2})}^{n} (1 - \frac{1}{x} + {\frac{1}{x^{2}}}^{n})$

Now, RHS $= {(1 + x + x^{2})}^{n} (1 - \frac{1}{x} + {\frac{1}{x^{2}}}^{n})$

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

Thus, $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + \dots + a_{2 n}^{2}$

$=$ Coefficient of the term independent of $x$ in

$\frac{1}{x^{2 n}} {(1 + x^{2} + x^{4})}^{n}$

$=$ Coefficient of $x^{2 n}$ in ${(1 + x^{2} + x^{4})}^{n}$

$=$ Coefficient of $t^{n}$ in ${(1 + t + t^{2})}^{n} = a_{n}$

Binomial Theorem 2 Question 13

16.

Solution:

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Binomial Theorem 2 Question 13

16.

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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