Question

Para

-1 \lt r \lt 1,

sea

S(r)

la suma de la serie geométrica

12 + 12r + 12r^2 + 12r^3 + \cdots.

Sea

a

entre

-1

y

1

que satisface

S(a)S(-a) = 2016.

Halla

S(a) + S(-a).

For

-1 \lt r \lt 1,

let

S(r)

denote the sum of the geometric series

12 + 12r + 12r^2 + 12r^3 + \cdots.

Let

a

between

-1

and

1

satisfy

S(a)S(-a) = 2016.

Find

S(a) + S(-a).

Respuesta

Cada serie es geométrica:

S(r) = \frac{12}{1 - r}

Each series is geometric:

S(r) = \frac{12}{1 - r}

Tanto

S(a)S(-a)

como

S(a) + S(-a)

dependen solo de

1 - a^2,

así que nunca necesitas hallar

a

Both

S(a)S(-a)

and

S(a) + S(-a)

depend only on

1 - a^2,

so you never need to find

a

La serie geométrica suma

S(r) = \frac{12}{1 - r}.

Por lo tanto

\begin{aligned} S(a)S(-a) &= \frac{12}{1 - a} \cdot \frac{12}{1 + a} \\ &= \frac{144}{1 - a^2} = 2016, \end{aligned}

así que

\frac{1}{1 - a^2} = 14.

Sumando las dos sumas sobre un común denominador,

\begin{aligned} S(a) + S(-a) &= \frac{12}{1 - a} \\ &\quad {}+ \frac{12}{1 + a} \\ &= \frac{24}{1 - a^2} \\ &= 24 \cdot 14 = 336. \end{aligned}

The geometric series sums to

S(r) = \frac{12}{1 - r}.

Therefore

\begin{aligned} S(a)S(-a) &= \frac{12}{1 - a} \cdot \frac{12}{1 + a} \\ &= \frac{144}{1 - a^2} = 2016, \end{aligned}

so

\frac{1}{1 - a^2} = 14.

Adding the two sums over a common denominator,

\begin{aligned} S(a) + S(-a) &= \frac{12}{1 - a} \\ &\quad {}+ \frac{12}{1 + a} \\ &= \frac{24}{1 - a^2} \\ &= 24 \cdot 14 = 336. \end{aligned}