VRAE oor vraestel 1: Sigma notasie met 2 sigmas

Hi Reynard

Awesome som!
Onthou 'n sigma notasie is 'n som op sy eie, so hier wil hulle hê jy moet twee aparte somme uitwerk en hulle dan gelyk stel aan mekaar.

Hier is my oplossing - jy kan hier click om 'die video oor hierdie som te kyk.

$table row cell sum from k equals 1 to infinity of 27 p to the power of k end cell row cell T subscript 1 equals 27 p to the power of 1 end cell row cell T subscript 2 equals 27 p squared end cell row cell T subscript 3 equals 27 p cubed end cell row blank row cell a equals 27 p end cell row cell r equals T subscript 2 over T subscript 1 equals fraction numerator 27 p squared over denominator 27 p end fraction equals p end cell row cell S subscript infinity equals ? end cell row blank row cell S subscript infinity equals fraction numerator a over denominator 1 minus r end fraction end cell row cell equals fraction numerator 27 p over denominator 1 minus p end fraction end cell end table$

$table row cell sum from t equals 1 to 12 of open parentheses 24 minus 3 t close parentheses end cell row blank row cell T subscript 1 equals 24 minus 3 open parentheses 1 close parentheses equals 21 end cell row cell T subscript 2 equals 24 minus 3 open parentheses 2 close parentheses equals 18 end cell row cell T subscript 3 equals 24 minus 3 open parentheses 3 close parentheses equals 15 end cell row blank row cell a equals 21 end cell row cell d equals T subscript 2 minus T subscript 1 equals 18 minus 21 equals negative 3 end cell row cell n equals 12 minus 1 plus 1 equals 12 end cell row cell S subscript 12 equals ? end cell row blank row cell S subscript n equals n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets end cell row cell S subscript 12 equals 12 over 2 open square brackets 2 open parentheses 21 close parentheses minus open parentheses 12 minus 1 close parentheses open parentheses negative 3 close parentheses close square brackets end cell row cell equals 54 end cell end table$

$table row cell sum from k equals 1 to infinity of 27 p to the power of k equals sum from t equals 1 to 12 of open parentheses 24 minus 3 t close parentheses end cell row cell fraction numerator 27 p over denominator 1 minus p end fraction equals 54 end cell row cell 27 p equals 54 open parentheses 1 minus p close parentheses end cell row cell 27 p equals 54 minus 54 p end cell row cell 81 p equals 54 end cell row cell p equals 2 over 3 end cell end table$