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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Similar questions

t^{8}

{(1 + t)}^{2} {(1 + t + t^{2} + . . . . + t^{9})}^{3}

t^{8}

{(1 + t)}^{2} {(1 + t + t^{2} + . . . . + t^{9})}^{3}

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

\frac{d^{2} y}{d x^{2}}

t = 0

t^{2} + t + 1 = 0,

(t + \frac{1}{t}) + (t^{2} + \frac{1}{t^{2}}) + \dots + (t^{27} + \frac{1}{t^{^{27}}})

t^{2} + t + 1 = 0

{(t + \frac{1}{t})}^{2} + {(t^{2} + \frac{1}{t^{2}})}^{2} + . . . . . . . . + {(t^{27} + \frac{1}{t^{27}})}^{2}

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