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# Evaluate the given expression $\sum\limits_{k = 1}^n {\left( {{2^k} + {3^{k - 1}}} \right)}$

Last updated date: 23rd May 2024
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Hint: Simplify the required expression by breaking the submission down into two different terms (as they can be simplified around the positive sign) then , expand by writing different values of k starting from 1, to obtain a progression series.

Now we need to find the value of $\sum\limits_{k = 1}^n {\left( {{2^k} + {3^{k - 1}}} \right)}$
So we can simplify it to
$\Rightarrow \sum\limits_{k = 1}^n {{2^k}} + \sum\limits_{k = 1}^n {{3^{k - 1}}}$

Now let’s expand each of these submissions by putting various values of k starting from 1 and going till n.

$\Rightarrow \left( {{2^1} + {2^2} + {2^3} + {2^4}............ + {2^n}} \right) + \left( {{3^0} + {3^1} + {3^2} + {3^3}............ + {3^{n - 1}}} \right)$ (EQ 1)
Now, let’s talk about the first expression which is
$\Rightarrow \left( {{2^1} + {2^2} + {2^3} + {2^4}............ + {2^n}} \right)$

Now, the first term of this series as ${a_1} = {2^1} = 2$ and second term ${a_2} = {2^2} = 4$
If we divide the second term by first term then we get common ratio ${r_1} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{4}{2} = 2$ (EQ 2)

Now, the third term of this series is${a_3} = {2^3} = 8$.
If we divide the fourth term with third term then we get a common ratio ${r_2} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{16}}{8} = 2$ (EQ 3)

Now, clearly equation (EQ 2) is equal to equation (EQ 3) so we can say that ${r_1} = {r_2} = 2$, hence the given series is in GP, because a series is in GP if and only if the common ratio remains constant throughout.

Now, this series is forming a sum of n terms where the common ratio is greater than 1.
So ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}{\text{ if r > 1}}$ (EQ 4)
Putting the values in equation (4) we get
${S_n} = \dfrac{{2\left( {{2^n} - 1} \right)}}{{2 - 1}} = \dfrac{{2\left( {{2^n} - 1} \right)}}{1} = 2 \times {2^n} - 2 = {2^{n + 1}} - 2$ (EQ 5)

Now, similarly talking about the second series
$\left( {{3^0} + {3^1} + {3^2} + {3^3}............ + {3^{n - 1}}} \right)$
Now, the first term of this series as${a_1} = {3^0} = 1$ and second term ${a_2} = {3^1} = 3$
If we divide the second term by first term then we get common ratio ${r_1} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{3}{1} = 3$ (EQ 6)

Now, the third term of this series is${a_3} = {3^2} = 9$.
If we divide the fourth term with third term then we get a common ratio ${r_2} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{9}{3} = 3$ (EQ 7)

Now, clearly equation (6) is equal to equation (7) so we can say that${r_1} = {r_2} = 3$, hence the given series is in GP, because a series is in GP if and only if the common ratio remains constant throughout.
Putting the values in equation (EQ 4) we get
${S_n} = \dfrac{{{3^0}\left( {{3^n} - 1} \right)}}{{3 - 1}} = \dfrac{{\left( {{3^n} - 1} \right)}}{2}$ (EQ 8)

Thus equation (1) is equal to equation (5) + equation (8)
$\Rightarrow \left( {{2^1} + {2^2} + {2^3} + {2^4}............ + {2^n}} \right) + \left( {{3^0} + {3^1} + {3^2} + {3^3}............ + {3^{n - 1}}} \right)$
$\Rightarrow {2^{n + 1}} - 2 + \dfrac{{\left( {{3^n} - 1} \right)}}{2}$
$\Rightarrow {2^{n + 1}} - 2 + \dfrac{{{3^n}}}{2} - \dfrac{1}{2} \\ \Rightarrow {2^{n + 1}} + \dfrac{{{3^n}}}{2} - \dfrac{5}{2} \\$

Note: Whenever we come across such types of problems the key concept that we need to recall is that a series is in GP if and only if the common ratio is coming out to be constant, moreover direct expansion of submissions always helps in making a series whose sum can be found using respective series formula.