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7.Binomial Theorem
hard
સંખ્યા $111......1$ ($91$ વખત) એ . . .
A
અવિભાજ્ય નથી.
B
યુગ્મ સંખ્યા
C
અયુગ્મ સંખ્યા નથી
D
એકપણ નહિ.
Solution
(a) $111…..1$ ($91$ times)
= $1 + 10 + {10^2} + ….. + {10^{90}}$
= $\frac{{{{10}^{91}} – 1}}{{10 – 1}} = \frac{{{{({{10}^7})}^{13}} – 1}}{{10 – 1}}$= $\frac{{{t^{13}} – 1}}{9}$, where $t = {10^7}$
= $\left( {\frac{{t – 1}}{9}} \right)\,({t^{12}} + {t^{11}} + ….. + t + 1)$
= $\left( {\frac{{{{10}^7} – 1}}{{10 – 1}}} \right)\,(1 + t + {t^2} + …. + {t^{12}})$
$ = (1 + 10 + {10^2} + …. + {10^6})(1 + t + {t^2} + … + {t^{12}})$
$111…..1(91\,\,{\rm{times)}}$ is a composite number.
Standard 11
Mathematics
