Question

Question only for Brainly Star and their team!


find sum of n terms of series...

5/6 + 7/ 54 + 9/324 +..........​

Answer 1

Answer:

a=5/6

d=7/54-9/6=-74/54

sum of n terms=n/2(2a+(n-1)d)

n/2(2*5/6+(n-1)-74/54)

n/2(5/3+74/54-74n/54)

n/2(164/54-74n/54)

n/108(164-74n)

164n-7992n²/108

hope it helps

Step-by-step explanation:

Answer 2

Answer:

$\large\bold\red{ S_{n} = 1 - \frac{1}{ {3}^{n}(n + 1) } }$

Step-by-step explanation:

Given,

a series,

$\frac{5}{6} + \frac{7}{54} + \frac{9}{324} + ...........$

Further,

we can simplify the Given series as,

$(\frac{5}{1 \times 2} \times \frac{1}{3} ) + (\frac{7}{2 \times 3} \times \frac{1}{ {3}^{2} } ) + ( \frac{9}{3 \times 4} \times \frac{1}{ {3}^{3} } )+ ........$

Here,

after observing the series,

we get,

the nth term of the series ,

$U_{n} = \frac{2n + 3}{n(n + 1)} \times \frac{1}{ {3}^{n} }$

Now,

Let's assume that,

$\frac{2n + 3}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$

Now,

comparing the coefficients of the like terms,

we get,

$= > A = 3 \\ \\ and \\ \\ = > B = - 1$

Therefore,

we have,

$= > U_{n} = ( \frac{3}{n} - \frac{1}{n + 1} ) \frac{1}{ {3}^{n} } \\ \\ = > U_{n} = ( \frac{1}{n} \times \frac{1}{ {3}^{n - 1} }) - (\frac{1}{n + 1} \times \frac{1}{ {3}^{n} } )$

Hence,

the sum of the sries,

$= > S_{n} = 1 - \frac{1}{n + 1} \times \frac{1}{ {3}^{n} } \\ \\ \bold{= > S_{n} = 1 - \frac{1}{ {3}^{n}(n + 1) } }$