7.Binomial Theorem
hard

$\left( \begin{array}{l}30\\0\end{array} \right)\,\left( \begin{array}{l}30\\10\end{array} \right) - \left( \begin{array}{l}30\\1\end{array} \right)\,\left( \begin{array}{l}30\\11\end{array} \right)$ + $\left( \begin{array}{l}30\\2\end{array} \right)\,\left( \begin{array}{l}30\\12\end{array} \right) + ....... + \left( \begin{array}{l}30\\20\end{array} \right)\,\left( \begin{array}{l}30\\30\end{array} \right)$ का मान है

A

$^{60}{C_{20}}$

B

$^{30}{C_{10}}$

C

$^{60}{C_{30}}$

D

$^{40}{C_{30}}$

(IIT-2005)

Solution

${(1 – x)^{30}} = {\,^{30}}{C_0}{x^0} – {\,^{30}}{C_1}{x^1} + {\,^{30}}{C_2}{x^2}$

$ + …… + {( – 1)^{30}}{\;^{30}}{C_{30}}{x^{30}}$….$(i)$

${(x + 1)^{30}} = {\,^{30}}{C_0}{x^{30}} + {\,^{30}}{C_1}{x^{29}} + {\,^{30}}{C_2}{x^{28}}$

  $ + …… + {\,^{30}}{C_{10}}{x^{20}} + …. + {\,^{30}}{C_{30}}{x^0}$….$(ii)$

(i) एवं (ii) का गुणा करके दोनों पक्षों $x^{20}$ के गुणांकों की तुलना करने पर अभीष्ट योग=  $x^{20}$ मे $(1 -x^2)^{30}$=${^{30}}{C^{10}}$.

Standard 11
Mathematics

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