(1-2 sqrt{x})^{50} के द्विपद प्रसार में x की | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${\left( {1 - 2\sqrt x } \right)^{50}} = {^{50}}{c_0} - {^{50}}{c_1}{\left( {2\sqrt x } \right)^1}$${\left( {1 - 2\sqrt x } \right)^{50}} = {^{50}}{c_

Vedclass · Accepted Answer

$\;\frac{1}{2}\left( {{3^{50}} + 1} \right)$

Vedclass · Answer

${\left( {1 - 2\sqrt x } \right)^{50}} = {^{50}}{c_0} - {^{50}}{c_1}{\left( {2\sqrt x } \right)^1}$${\left( {1 - 2\sqrt x } \right)^{50}} = {^{50}}{c_0} - {^{50}}{c_1}{\left( {2\sqrt x } \right)^1}$$ + {^{50}}{c_4}\left( {2\sqrt x } \right)4.....$

$s = {^{50}}{c_0} + {^{50}}{c_2} \cdot {2^2} + {^{50}}{c_4}{\left( 2 \right)^4} + .... + {^{50}}c_{502}^{50}$

${\left( {1 + x} \right)^{50}} = 1 + {^{50}}{C_1}{x^1} + {^{50}}{c_2}{x^2} + ...$

$x = 2, - 2$

${\left( {1 + 2} \right)^{50}} = 1 + {^{50}}{c_1}\left( 2 \right) + {^{50}}{c_2}{\left( 2 \right)^2} + ..$

${\left( {1 - 2} \right)^{50}} = 1 - {^{50}}{c_1}\left( 2 \right) + $${^{50}}{c_2}{\left( 2 \right)^2} - {^{50}}{C_3}{\left( 2 \right)^3} + .....$

${3^{50}} + 1$$ = 2\left[ {{^{50}}{C_0} + {^{50}}{c_2}{{\left( 2 \right)}^2} + {^{50}}{c_4}{{\left( 2 \right)}^4} + ..} \right]$

$\frac{{{3^{50}} + 1}}{2}$$ = {^{50}}{C_0} + {^{50}}{c_2}\left( 2 \right) + {^{50}}{c_4}{\left( 2 \right)^4} + ...$

$(1-2 \sqrt{x})^{50}$ के द्विपद प्रसार में $x$ की पूर्णांकीय घातों के गुणांकों का योग है

Similar Questions

$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $

यदि ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^n},$ तो $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

व्यंजक $(5+x)^{500}+x(5+x)^{499}+x^2(5+x)^{498}+\ldots . x^{500}$ $x > 0$ में $x ^{101}$ का गुणांक होगा -

यदि $n$ एक धनात्मक पूर्णांक है तथा ${C_k} = {\,^n}{C_k}$, तब ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =

यदि ${a_k} = \frac{1}{{k(k + 1)}},$ जबकि $k = 1,\,2,\,3,\,4,.....,\,n$, तब ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $