If $f (x) = 1 + x + x^{2} + . . . . . + x^{100}$ then find $f^{'} (1)$ .

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Solution

We need to find the derivative of $f (x)$ at $x = 1$
i.e. $f^{'} (1)$
$f (x) = 1 + x + x^{2} + . . . . . . + x^{100}$
$\Rightarrow f^{'} (x) = (1 + x + x^{2} + . . . . . + x^{100})^{'}$
$\Rightarrow f^{'} (x) = 0 + 1 + 2 x + . . . . . + 100 x^{99}$
$∴ f^{'} (x) = 1 + 2 x + . . . . . + 100 x^{99}$
Putting $x = 1$ in $f^{'} (x)$
$\Rightarrow f^{'} (1) = 1 + 2 \times 1 + . . . . . + 100 \times 1^{99}$
$\Rightarrow f^{'} (1) = 1 + 2 + . . . . + 100$
$\Rightarrow f^{'} (1) = \frac{100 (100 + 1)}{2}$ = $5050$
$\Rightarrow f^{'} (1)$ = $5050$

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