Question

sum of n terms of the series 1/2+3/4+7/8+15/16+ . . . is equal to

Answer 1

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Back to the ¿question? of yours human,

So the numbers we have here are,

$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} .....$

This could also be written as,

$(1 - \frac{1}{2} ) + (1 - \frac{1}{4} ) + (1 - \frac{1}{8} ) + (1 - \frac{1}{16} )....$

Which is the same thing as,

$(1 - \frac{1}{2} ) + (1 - \frac{1}{ {2}^{2} }) + (1 - \frac{1}{ {2}^{3} } )...(1 - \frac{1}{ {2}^{n} } )$

Taking "n" times 1 common,

$1 + 1 + 1...n - ( \frac{1}{2} + \frac{1}{ {2}^{2} } + .. \frac{1}{ {2}^{n} } )$

As "n" times one is same thing as "n"

Therefore we have,

$n - ( \frac{1}{2} + \frac{1}{ {2}^{2} } ....+ \frac{1}{ {2}^{n} } )$
Take this ⬆️ as equation (i)

As you can see, the part after minus sign is in G.P. as the ratio of the consecutive terms is constant as 1/2.

You might know that,
when terms are in G.P. the sum of "n" terms equals,
$\frac{rl - a}{r - 1}$

Where:

"r" is the common ratio
"l" is the last term
and
"a" is the first term,

Therefore,
In our case it equals,
$\frac{ \frac{1}{2} ( \frac{1}{ {2}^{n} } ) - \frac{1}{2} }{ \frac{1}{2} - 1} \\ \\ \\ = \frac{ \frac{1}{2}( \frac{1}{ {2}^{n} } - 1 )}{ - \frac{1}{2} } \\ \\ \\ = -( \frac{1}{ {2}^{n} } - 1 ) \\ \\ = 1 - \frac{1}{ {2}^{n} }$

Substituting this value for the G.P series in equation (i)

And You'll end up with,

$n - (1 - \frac{1}{ {2}^{n} } ) \\ = n - 1 + \frac{1}{ {2}^{n} }$

And

Cheers !!!
you have the answer.

Hope I didn't bored you users.

Answer 2

Answer:

The sum is $n-1+\frac{1}{2^n}$

Step-by-step explanation:

Given series is

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+----$

we have to find the sum of n terms.

The above series can be written as

$(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})+---n$

$(1-\frac{1}{2})+(1-\frac{1}{2^2})+(1-\frac{1}{2^3})+(1-\frac{1}{2^4})+---(1-\frac{1}{2^n})$

$(1+1+...n)-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....)$

$n-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....)$

The above is G.P  (sum of GP is $\frac{a(1-r^n)}{1-r}$)

$n-\frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2})}$

$n-(1-\frac{1}{2^n})$

$n-1+\frac{1}{2^n}$