Question

If $1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n} = \dfrac{{\left( {4n - 1} \right){5^{a + 1}} + b}}{{16}}$ , then
A. $a = n,b = 1$
B. $a = n,b = 5$
C. $a = 1,b = 5$
D. $a = n,b = n$

Answer 1

First, we shall analyze the given information so that we are able to solve the given problem. Here, we are given that $1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n} = \dfrac{{\left( {4n - 1} \right){5^{a + 1}} + b}}{{16}}$
We are asked to calculate the values of $a$ and $b$ . We need to consider ${S_n} = 1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n}$
Then we shall multiply it by five. An, we need to subtract both the resultant equations. Then, we will note that the sequence is in geometric progression. Then we need to apply the formula of the sum of ${n^{th}}$ terms of G.P
Formula to be used:
The formula to calculate the sum of ${n^{th}}$ terms of G.P is as follows.
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$ where $a$ is the first term of G.P, $r$ is the common ratio of G.P, and is $n$ the number of terms.

Complete step by step answer:
It is given that $1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n} = \dfrac{{\left( {4n - 1} \right){5^{a + 1}} + b}}{{16}}$
We are asked to calculate the values of $a$ and $b$ .
Let ${S_n} = 1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n}$ ……. $\left( 1 \right)$
We need to multiply the equation $\left( 1 \right)$ by $5$
Hence, we get
$5{S_n} = 1.5 \times 5 + {2.5^2} \times 5 + {3.5^3} \times 5 + .... + n{.5^n} \times 5$
$5{S_n} = {1.5^2} + {2.5^3} + {3.5^4} + .... + n{.5^{n + 1}}$ ………. $\left( 2 \right)$
We need to subtract the equation $\left( 1 \right)$ from the equation $\left( 2 \right)$ .
${S_n} - 5{S_n} = 1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n} - {1.5^2} - {2.5^3} - {3.5^4} - ....\left( {n - 1} \right){5^n} - n{.5^{n + 1}}$
$- 4{S_n} = 1.5 + {2.5^2} - {1.5^2} + {3.5^3} - {2.5^3} + .... + n{.5^n} - \left( {n - 1} \right){5^n} - n{.5^{n + 1}}$
$\Rightarrow - 4{S_n} = 1.5 + {1.5^2} + {1.5^3} + .... + {1.5^n} - n{.5^{n + 1}}$ ………… $\left( 3 \right)$
Here, we shall note that $1.5 + {1.5^2} + {1.5^3} + .... + {1.5^n}$ is in geometric progression.
$1.5 + {1.5^2} + {1.5^3} + .... + {1.5^n} = 1.5 + 1.5 \times 5 + 1.5 \times {5^2} + .... + 1.5 \times {5^{n - 1}}$
Hence the first term will be $a = 5$ and the common ratio will be $r = \dfrac{{{5^2}}}{5} = 5$
Now, we shall apply the formula of the sum of ${n^{th}}$ terms of G.P, ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$
$1.5 + {1.5^2} + {1.5^3} + .... + {1.5^n} = 1.5 + 1.5 \times 5 + 1.5 \times {5^2} + .... + 1.5 \times {5^{n - 1}}$
$= \dfrac{{5\left( {{5^n} - 1} \right)}}{{5 - 1}}$
$= \dfrac{{5\left( {{5^n} - 1} \right)}}{4}$
Now, we shall substitute the above result in the equation $\left( 3 \right)$ .
$\Rightarrow - 4{S_n} = 1.5 + {1.5^2} + {1.5^3} + .... + {1.5^n} - n{.5^{n + 1}}$
$\Rightarrow - 4{S_n} = \dfrac{{5\left( {{5^n} - 1} \right)}}{4} - n{.5^{n + 1}}$
$= \dfrac{{{5^{n + 1}} - 5}}{4} - n{.5^{n + 1}}$
$= \dfrac{{{5^{n + 1}} - 5 - 4n{{.5}^{n + 1}}}}{4}$
$\Rightarrow - 16{S_n} = {5^{n + 1}} - 4n{.5^{n + 1}} - 5$
$\Rightarrow - 16{S_n} = {5^{n + 1}}\left( {1 - 4n} \right) - 5$
Multiplying throughout by minus sign, we get
$\Rightarrow 16{S_n} = {5^{n + 1}}\left( {4n - 1} \right) + 5$
$\Rightarrow {S_n} = \dfrac{1}{{16}}\left[ {{5^{n + 1}}\left( {4n - 1} \right) + 5} \right]$
$\Rightarrow {S_n} = \dfrac{1}{{16}}\left[ {\left( {4n - 1} \right){5^{n + 1}} + 5} \right]$ …………. $\left( 4 \right)$
It is given that ${S_n} =$ $1.5 + {2.5^2} + {3.5^3} + .... + n{.5^n} = \dfrac{{\left( {4n - 1} \right){5^{a + 1}} + b}}{{16}}$ ………….. $\left( 5 \right)$
Now, we shall compare the equations $\left( 4 \right)$ and $\left( 5 \right)$
That is,
$\dfrac{1}{{16}}\left[ {\left( {4n - 1} \right){5^{n + 1}} + 5} \right] = \dfrac{{\left( {4n - 1} \right){5^{a + 1}} + b}}{{16}}$
Since all the terms are equal except $a$ and $b$ , we can get the answer $a = n,b = 5$

So, the correct answer is “Option B”.

Note: If we are given the sum of ${n^{th}}$ terms of G.P, then we need to apply the formula ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$ and ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$ where $a$ is the first term of G.P, $r$ is the common ratio of G.P and is $n$ the number of terms.
Since the common ratio is greater than one, we applied the formula ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$
Suppose the common ratio is smaller than one, we need to apply the formula ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$