Value of left( begin{array}{l}300end{array} right),left( begin{array}{l}3010end{array} right)

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7.Binomial Theorem

hard

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

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

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

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

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

(IIT-2005)

Solution

(b) ${(1 – x)^{30}} = {\,^{30}}{C_0}{x^0} – {\,^{30}}{C_1}{x^1} + {\,^{30}}{C_2}{x^2} + …… + {( – 1)^{30}}{\;^{30}}{C_{30}}{x^{30}}$….$(i)$

${(x + 1)^{30}} = {\,^{30}}{C_0}{x^{30}} + {\,^{30}}{C_1}{x^{29}} + {\,^{30}}{C_2}{x^{28}}+ …… + {\,^{30}}{C_{10}}{x^{20}} + …. + {\,^{30}}{C_{30}}{x^0}$….$(ii)$

Multiplying (i) and (ii) and equating the coefficient of $x^{20}$ on both sides, we get required sum

= coefficient of $x^{20}$ in $(1 -x^2)^{30}={^{30}}{C^{10}}$.

Standard 11

Mathematics

The coefficent of $x^7$ in the expansion of ${\left( {1 – x – {x^2} + {x^3}} \right)^6}$ is

hard

(AIEEE-2011)

$\frac{1}{{1!(n – 1)\,!}} + \frac{1}{{3!(n – 3)!}} + \frac{1}{{5!(n – 5)!}} + …. = $

medium

The number of terms in the expansion of $(1 +x)^{101} (1 +x^2 – x)^{100}$ in powers of $x$ is

hard

(JEE MAIN-2014)

$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered}
4 \hfill \\
n \hfill \\
\end{gathered} \right)} {\left( { – 1} \right)^n}$ is

normal

The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + ….. + (a + nb)C_n$ is where $Cr's$ denotes combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$

normal

Solution

Similar Questions

The coefficent of $x^7$ in the expansion of ${\left( {1 – x – {x^2} + {x^3}} \right)^6}$ is

$\frac{1}{{1!(n – 1)\,!}} + \frac{1}{{3!(n – 3)!}} + \frac{1}{{5!(n – 5)!}} + …. = $

The number of terms in the expansion of $(1 +x)^{101} (1 +x^2 – x)^{100}$ in powers of $x$ is

$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered} 4 \hfill \\ n \hfill \\ \end{gathered} \right)} {\left( { – 1} \right)^n}$ is

The sum of the series $aC_0 + (a + b)C_1 + (a + 2b)C_2 + ….. + (a + nb)C_n$ is where $Cr's$ denotes combinatorial coefficient in the expansion of $(1 + x)^n, n \in N$

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$\sum\limits_{n = 0}^4 {{{\left( {1009 – 2n} \right)}^4}\left( \begin{gathered}
4 \hfill \\
n \hfill \\
\end{gathered} \right)} {\left( { – 1} \right)^n}$ is