# The recursive sequence is defined by the formula t_n=2t_(n-1)+3; and t_1=-2, how do you find t_6?

May 26, 2016

Find ${t}_{n} = {2}^{n - 1} - 3$ and hence ${t}_{6} = {2}^{5} - 3 = 29$

#### Explanation:

Note that as $n$ and ${t}_{n}$ get larger, the sequence will approximate a geometric sequence with common ratio $2$.

So we can look for a general formula for terms of the form:

${t}_{n} = {2}^{n} a + b$

for some constants $a$ and $b$.

Then we find:

${2}^{n} a + b = {t}_{n} = 2 {t}_{n - 1} + 3 = 2 \left({2}^{n - 1} a + b\right) = {2}^{n} a + 2 b + 3$

Hence we find $b = - 3$

Then:

$- 2 = {t}_{1} = {2}^{1} a - 3 = 2 a - 3$

Hence we find $a = \frac{1}{2}$

So the general formula of a term of our sequence is:

${t}_{n} = {2}^{n - 1} - 3$

If you like, we can double check this formula:

$2 {t}_{n - 1} + 3 = 2 \left({2}^{n - 2} - 3\right) + 3 = {2}^{n - 1} - 6 + 3 = {2}^{n - 1} - 3 = {t}_{n}$

In particular:

${t}_{6} = {2}^{5} - 3 = 32 - 3 = 29$

May 26, 2016

${t}_{6} = {2}^{5} - 3$

#### Explanation:

Proposing ${t}_{n} = {c}_{0} {a}^{n} + b$ and substituting we get
${c}_{0} {a}^{n} + b = 2 \left({c}_{0} {a}^{n - 1} + b\right) + 3$ or
${c}_{0} {a}^{n} - 2 {c}_{0} {a}^{n - 1} = 2 b - b + 3$.

Making $a = 2$ and $b = - 3$ the equality is observed. So we get

${t}_{n} = {c}_{0} {2}^{n} - 3$ and also
${t}_{1} = {c}_{0} {2}^{1} - 3 \to {c}_{0} = \frac{1}{2}$. Putting all together
${t}_{n} = {2}^{n - 1} - 3 \to {t}_{6} = {2}^{5} - 3$