Question

the value of 8^1/2×8^1/4×8.....infinity​

Answer 1

GIVEN :–

$\\ \bf \implies {8}^{ \frac{1}{2}} \times {8}^{ \frac{1}{4}} \times {8}^{ \frac{1}{8} } \times .......... \infty$

TO FIND :–

• Value = ?

SOLUTION :–

• Let the series –

$\\ \bf \implies p = {8}^{ \frac{1}{2}} \times {8}^{ \frac{1}{4}} \times {8}^{ \frac{1}{8} } \times .......... \infty$

• Using identity –

$\bf \implies {a}^{b} \times{a}^{c} = {a}^{b + c}$

• So that –

$\\ \bf \implies p = {8}^{ \left\{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....... \infty\right \}}$

• We also know that sum of infinite terms of G.P. –

$\\ \implies \large{ \boxed{ \bf s = \dfrac{a}{1 - r}}}$

• Here –

$\\ \bf \: \: {\huge{.}} \: \: a = \dfrac{1}{2}$

$\\ \bf \: \: {\huge{.}} \: \: d= \dfrac{1}{2}$

• So that –

$\\ \bf \implies p = {8}^{\left\{ \dfrac{ \frac{1}{2} }{1 - \frac{1}{2} } \right\}}$

$\\ \bf \implies p = {8}^{\left\{ \cancel\dfrac{ \frac{1}{2} }{\frac{1}{2} } \right\}}$

$\\ \bf \implies p = {8}^{\left(1\right)}$

$\\ \large \implies{ \boxed{ \bf p =8}}$

Answer 2

Given :-

• ${8}^{ \frac{1}{2} } \times {8}^{ \frac{1}{4} } \times {8}^{ \frac{1}{8} } - - - - \infty$

Solution :-

Let the series →

$: \implies \: p = {8}^{ \frac{1}{2} } \times {8}^{ \frac{1}{4} } \times {8}^{ \frac{1}{8} } - - - - \infty$

We know that :-

• ${a}^{m} \times {a}^{n} = {a}^{m + n}$

So,

$: \implies \: p = {8}^{ \big(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} - - - \infty \big)}$

Now,

The sum of infinity terms in G.P. -

$s \: = \frac{a}{1 - d}$

Where 'a' is the first term and 'd' is the difference between each term.

So,

here ,

$\star \: a \: = \frac{1}{2} \\ \\ \star \: d = \frac{1}{2}$

$: \implies \: p = {8}^{ \big(\frac{ \frac{1}{2} }{1 - \frac{1}{2} } \big) } \\ \\ : \implies \: p = {8}^{ \big( \frac{ \frac{1}{2} }{ \frac{2 - 1}{2} } \big)} \\ \\ : \implies \: p = {8}^{ \big( \frac{ \frac{1}{2} }{ \frac{1}{2} } \big)} \\ \\ : \implies \: p = {8}^{1} \\ \\ : \implies \boxed{ p = 8}$

Hence, the value of given series is 8.