If Σ_{ r =0}^5 frac{{ }^{11} C _{2 r +1}}{2 r +2}=frac{ m }{ n }, operatorname{gcd}( m , n )=1 , th

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7.Binomial Theorem

hard

If $\sum_{ r =0}^5 \frac{{ }^{11} C _{2 r +1}}{2 r +2}=\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m - n$ is equal to _____

A$2785$

B$2035$

C$5039$

D$2235$

(JEE MAIN-2025)

Solution

$\int_0^1(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_0^1$
$\frac{2^{12}-1}{12}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+\frac{C_3}{4}+\ldots$
$\int_{-1}^0(1+x)^{11} d x=\left[C_0 x+\frac{C_1 x^2}{2}+\frac{C_2 x^3}{3}+\ldots\right]_{-1}^0$
$\frac{1}{12}=C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\ldots$
$\frac{2^{12}-2}{12}=2\left(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots\right)$
$\frac{C_1}{2}+\frac{C_3}{4}-\frac{C_5}{6}+\ldots=\frac{2^{11}-1}{12}=\frac{2047}{12}$

Standard 11

Mathematics

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is …. .

medium

(JEE MAIN-2021)

The sum of last eigth coefficients in the expansion of $(1 + x)^{15}$ is :-

normal

$\left( {\begin{array}{{20}{c}}n\\0\end{array}} \right) + 2\,\left( {\begin{array}{{20}{c}}n\\1\end{array}} \right) + {2^2}\left( {\begin{array}{{20}{c}}n\\2\end{array}} \right) + ….. + {2^n}\left( {\begin{array}{{20}{c}}n\\n\end{array}} \right)$ is equal to

medium

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

hard

(IIT-1975)

$\frac{{{C_0}}}{1} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + …. + \frac{{{C_n}}}{{n + 1}} = $

medium

If $\sum_{ r =0}^5 \frac{{ }^{11} C _{2 r +1}}{2 r +2}=\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m - n$ is equal to _____

Solution

Similar Questions

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is …. .

The sum of last eigth coefficients in the expansion of $(1 + x)^{15}$ is :-

$\left( {\begin{array}{*{20}{c}}n\\0\end{array}} \right) + 2\,\left( {\begin{array}{*{20}{c}}n\\1\end{array}} \right) + {2^2}\left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + ….. + {2^n}\left( {\begin{array}{*{20}{c}}n\\n\end{array}} \right)$ is equal to

If ${a_1},{a_2},{a_3},{a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$, then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

$\frac{{{C_0}}}{1} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + …. + \frac{{{C_n}}}{{n + 1}} = $

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$\left( {\begin{array}{{20}{c}}n\\0\end{array}} \right) + 2\,\left( {\begin{array}{{20}{c}}n\\1\end{array}} \right) + {2^2}\left( {\begin{array}{{20}{c}}n\\2\end{array}} \right) + ….. + {2^n}\left( {\begin{array}{{20}{c}}n\\n\end{array}} \right)$ is equal to