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  4. The sum of the coefficients in the expansion of (1+x-3 x2)3148 is

Q. The sum of the coefficients in the expansion of $\left(1+x-3 x^{2}\right)^{3148}$ is

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Solution:

$\left(1+x-3 x^{2}\right)^{3148}=A o+A_{1} x+A_{2} x^{2}+\ldots$
Put $x=1 (1+1-3)^{3148}=A 0+A_{1}+A_{2}+\ldots$
$(-1)^{3148}=A_{0}+A_{1}+A_{2}+\ldots$
$\therefore A_{0}+A_{1}+A_{2}+\ldots=1$

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