#### Answer

$100y^{99}$

#### Work Step by Step

We find the 100th term in the expansion of $(1+y)^{100}$ by using Binomial Theorem coefficients:
$\left(\begin{array}{l}
100\\99\end{array}\right)1^{1}y^{99}=100(1^1)(y^{99})=100y^{99}$

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$100y^{99}$

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