Question

Find the $8^{th}$ term of the series 1 + 5 + 18 + 58 + 179 + ..........

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Answer 1

We see that squence of first consecutive difference 4,13,40,121...........and second consecutive difference 9,27,81......... which is a GP.Let $T_n$ be the $n^{th}$ term of the series.

Let
$S_n$ = 1 + 5 + 18 + 58 + 179 + ................... $T_n$----------------------(1)

$S_n$ = 1 + 5 + 18 + 58 + ................... $T_n-1$ + $T_n$----------------------(2)

Write the $S_n$ in such a way that $1^{th}$ term of equation 2 comes under $2^{nd}$ term of equation 1 and so on.......

Substrating equation 2 from equation 1.

0 = 1 + 4 + 13 + 40 + 121 + ................... ($T_n$ - $T_{n-1}$) - $T_n$

Let $t_n$ is the $n^{th}$ term of the below given series

$T_n$ = 1 + 4 + 13 + 40 + 121 + ................... $t_n$---------------(3)

$T_n$ = 1 + 4 + 13 + 40+ ................... ($t_n-1$) + $t_n$------------------(4)

Write the $T_n$ in such a way that $1^{th}$ term of equation 4 comes under $2^{nd}$ term of equation 3 and so on........

Substrating equation 4 from equation 3.

0 = 1 + 3 + 9 + 27 + 81 + ................... ($t_n$ - $t_{n-1}$) - $t_n$

$t_n$ = 1 + 3 + 9 + 27 + 81 + ...................n Terms

$t_n$ is a GP

Sum of the geometric progression

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

Since $T_n=\sum t_n$

= $\frac{1}{2}\sum (3^n-1)$

=$\frac{1}{2}\sum \left(3^n\right)-\sum 1$

= $\frac{1}{2}$[($3+3^2+3^3+3^4+............nterms$) - n]

=$\frac{1}{2}$ $[\frac{3(3^n-1)}{3-1}-n]$

=$\frac{3}{4}$ ($3^n$ - 1) - $\frac{n}{2}$ ------------------(5)

Put n=8 in equation 5 for ${10}^{th}$ term of the series in equation 1

$T_{10}$ = $\frac{3}{4}$ ($3^8$ - 1) - $\frac{8}{2}$ = 4916