The coefficient of $x^{50}$ in ${(1 + x)}^{41} {(1 - x + x^{2})}^{40}$ is ...........

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Solution

$f (x) = {(1 + x)}^{41} \times {(1 - x + x^{2})}^{40}$

$f (x) = {(1 + x)}^{1} \times {((1 + x) (1 - x + x^{2}))}^{40}$

$f (x) = {(1 + x)}^{1} \times {(1 - x + x^{2} + x - x^{2} + x^{3})}^{40}$

$f (x) = {(1 + x)}^{1} \times {(1 + x^{3})}^{40}$

$f (x) = {(1 + x^{3})}^{40} + x {(1 + x^{3})}^{40}$

All the terms in the expansion ${(1 + x^{3})}^{40}$ is in the form $x^{3 r}$ and in the $x {(1 + x^{3})}^{40}$ is of the form $x^{3 p + 1}$

So $3 r = 50$ or $3 p + 1 = 50$ Where we do not get integer values of $r, p$ so the coefficient of $x^{50}$ is $0$

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