(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + | vedclass

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Question

As we are interested in coefficient of $t^{50},$ we shall ignore all the term with exponent more than $50 .$ Thus we can write as 

$\left( {1

Vedclass · Accepted Answer

$1 + ^{25}C_5$

Vedclass · Answer

As we are interested in coefficient of $t^{50},$ we shall ignore all the term with exponent more than $50 .$ Thus we can write as 

$\left( {1 + {\,^{25}}{{m{C}}_1}{{m{t}}^2} +  \ldots  \ldots  + {\,^{25}}{{m{C}}_{25}}{{m{t}}^{50}}} ight) 	imes \left( {1 + {{m{t}}^{25}} + {{m{t}}^{40}} + {{m{t}}^{45}} + {{m{t}}^{47}}} ight)$

As all the terms in the first have even exponent we can ignore $\mathrm{t}^{25}, \mathrm{t}^{45}$ and $\mathrm{t}^{47}$ too thus coefficient of $\mathrm{t}^{50}$ is $ = {\,^{25}}{{m{C}}_{25}} + {\,^{25}}{{m{C}}_5} = 1 + {\,^{25}}{{m{C}}_5}$

$(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ ના વિસ્તરણમાં $t^{50}$ નો સહગુણક મેળવો

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