Question

prove that 3^100+3^101+3^102 / 3^99+3^98-3^97 =351/11

Answer 1

$\frac{ {3}^{100} + {3}^{101} + {3}^{102} }{ {3}^{99} + {3}^{98} - {3}^{97} } \\ \\ \\ => \frac{ {3}^{100} (1 + 3 ^{1} + {3}^{2} )}{ {3}^{97} ( {3}^{2} + {3}^{1} - 1) } \\ \\ \\ => \frac{ {3}^{100 - 97}(1 + 3 + 9)}{(9 + 3 - 1) } \\ \\ \\ = > \frac{ {3}^{3}(13) }{11} \\ \\ \\ => \frac{27 \times 13}{11} \\ \\ = > \frac{351}{11}$



Hence, proved.

Answer 2

Answer:

351/11

Step-by-step explanation:

3^100(1+3^1+3^2)/3^97(3^2+3^1-1)=3^100-97(1+3+9)/(9+3-1)= 3^3(13)/11= 27×13/11=351/11

hence proved