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7.Binomial Theorem
normal
The coefficient of $x^{49}$ in the expansion of $(x - 1)$$\left( {x\, - \,\frac{1}{2}\,} \right)$$\left( {x\, - \,\frac{1}{{{2^2}}}\,} \right)$ .....$\left( {x\, - \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to
A
$-2 \left( {1\, - \,\frac{1}{{{2^{50}}}}\,} \right)$
B
$+$ ve coefficient of $x$
C
$-$ ve coefficient of $x$
D
$-2 \left( {1\, - \,\frac{1}{{{2^{49}}}}\,} \right)$
Solution
coeff.. of $x^{49}$ in this series is
$-\left[ {1 + \frac{1}{2}\, + \frac{1}{{{2^2}}}\, + …… + \frac{1}{{{2^{49}}}}} \right]$ $=$$ – \,\left[ {\frac{{1 – \frac{1}{{{2^{50}}}}}}{{1 – \frac{1}{2}}}} \right]$ $=$$ – \,2.\,\left[ {1 – \frac{1}{{{2^{50}}}}} \right]$
Standard 11
Mathematics
