The coefficient of $t^{8}$ in ${(1 + t)}^{2} {(1 + t + t^{2} + . . . . + t^{9})}^{3}$ is

144

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145

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146

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147

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Solution

The correct option is B $145$
Coefficient of $t^{8}$ in ${(1 + t)}^{2} {(1 + t + t^{2} + . . . . + t^{9})}^{3}$
$=$ Coefficient of $t^{8}$ in $(1 + 2 t + t^{2}) {(1 + t + t^{2} + . . . . + t^{9} + . . . . + \infty)}^{3}$
$=$ Coefficient of $t^{8}$ in $(1 + 2 t + t^{2}) {(1 - t)}^{- 3}$

We know that ${(1 - x)}^{- r} = \sum^{n + r - 1} C_{n} x^{n}$

Here, there are two brackets. If we take $t^{2}$ from the first bracket , we need $t^{6}$ from the second bracket. In this case the coefficient $= 1 \cdot^{8} C_{2}$

$2 t$ from the first bracket and $t^{7}$ from the second bracket, then the coefficient $= 2 \cdot^{9} C_{2}$

$1$ from the first bracket and $t^{8}$ from the second bracket, then the coeeficient $= 1 \cdot^{10} C_{2}$

$Total = 1 \cdot^{8} C_{2} + 2 \cdot^{9} C_{2} + 1 \cdot^{10} C_{2} = 145$

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Find the coefficient of $t^{8}$ in the expansion of $(1 + 2 t^{2} - t^{3})^{9}$ .

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