Question

If $n^{th}$ term of the series 5 + 7 + 13 + 31+ 85 ------------- can be written as $T_n$ = a.$3^{(n-1)}$ + bn + c. Find the sum of the first eight terms of the given series.

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Answer 1

$T_n$ = a.$3^{n-1}$ + bn + c------------------------(1)

putting n = 1,2,3,................

We get,

$T_1$ = a + b + c = 5-----------------(2)

$T_2$ = 3a + 2b + c = 7---------------(3)

$T_3$ = 9a + 3b + c = 13--------------(4)

Multiply 3 in equation 2 and subtrate equation 3 from it.

b + 2c = 8------------------(5)

Multiply 9 in equation 2 and subtrate equation 4 from it.

6b + 8c = 32------------------(6)

Solving equation 5 and equation 6

b = 0,c = 4

Substitute and c in equation 2.

a + b + c = 5

a + 0 + 4 = 5

a = 1

So,a=1, b=0 and c=4

Substituting a,b,c values in equation 1. we get,

$T_n$ = $3^{n-1}$ + 4

Sum of the given series

$S_n={\sum }_{j=1}^n{T}_n$
= ${\sum }_{i=1}^n{3}^{n-1}$ + ${\sum }_{i=1}^{n}4$

= $3^0+3^1+3^3+........................)$ + 4n

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

$S_n$ = $\frac{3^n-1}{2}$ + 4n------------------------(7)

To get the sum of 8 terms

Substitute n = 8 in equation 7

$S_8$ = $\frac{3^8-1}{2}$ + $4\times 8$

= 3312