Question

Consider the following series in which if $1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}=\dfrac{\left( 2n-1 \right){{3}^{a}}+b}{4}$ then find the values of a & b.

Answer 1

Hint: First of all assume that the summation of the series which is on the left hand side of the equation is “S” so equate the above series to “S” then multiply left hand and right hand side by 3 then subtract this new series after multiplication by 3 from the original series. After subtraction, you will find a new series which is a G.P. and we know the formula for the summation of a series which is in a G.P.

Complete step-by-step answer:
The summation series given in the question is:
$1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}$
Now, let us assume this series as “S” so equating the above series to S.
$S=1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}$
Multiplying 3 on both the sides of the equation we get,
$S=1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}$
$\Rightarrow 3S={{3}^{2}}+{{2.3}^{3}}+{{3.3}^{4}}+.....+\left( n-1 \right){{.3}^{n}}+n{{.3}^{n+1}}$
Now, if you align numbers of the second series with the first series in such a way that ${{3}^{2}}$ of the second series lies just below ${{2.3}^{2}}$ of the first series and write ${{2.3}^{3}}$ of the second series just below ${{3.3}^{3}}$ of the first series. Similarly, you can write other elements of the series also. When you align the series in such a manner then subtraction of the two series will be quite easy. Subtracting the above series in a manner that we have just described we get,
$\begin{aligned} & \text{ }S=1.3+{{2.3}^{2}}+{{3.3}^{3}}+.......+n{{.3}^{n}} \\\ & \dfrac{-3S=\text{ }{{1.3}^{2}}+{{2.3}^{3}}+.......+\left( n-1 \right){{3}^{n}}+n{{.3}^{n+1}}}{-2S=3+{{3}^{2}}+{{3}^{3}}+........................+{{3}^{n}}-n{{.3}^{n+1}}} \\\ \end{aligned}$
Solving the above equation by dividing -2 on both the sides we get,
$S=-\dfrac{1}{2}\left( 3+{{3}^{2}}+{{3}^{3}}+.....+{{3}^{n}}-n{{.3}^{n+1}} \right)$………… Eq. (1)
As you can see from the above equation that a G.P. is forming inside the bracket except the term $-n{{.3}^{n+1}}$ so we are going to apply the summation of a G.P. in the series.
$3+{{3}^{2}}+{{3}^{3}}+.........+{{3}^{n}}$
Let us assume that the summation of the G.P. is ${{S}_{n}}$ so substituting the summation of the series as ${{S}_{n}}$ in eq. (1) we get,
$S=-\dfrac{1}{2}\left( {{S}_{n}}-n{{.3}^{n+1}} \right)$…….. Eq. (2)
We know that summation of a G.P. with first term “a” and the common ratio is “r” is:
${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ ………. Eq. (3)
Using the above formula in finding the sum of the below series:
$3+{{3}^{2}}+{{3}^{3}}+.........+{{3}^{n}}$
The above series has first term “a” is equal to 3 and the common ratio “r” of the G.P. is 3 so substituting these values in eq. (3) we get,
$\begin{aligned} & {{S}_{n}}=\dfrac{3\left( {{3}^{n}}-1 \right)}{3-1} \\\ & \Rightarrow {{S}_{n}}=\dfrac{{{3}^{n+1}}-3}{2} \\\ \end{aligned}$
Substituting the above value in eq. (2) we get,
$\begin{aligned} & S=-\dfrac{1}{2}\left( \dfrac{{{3}^{n+1}}-3}{2}-n{{.3}^{n+1}} \right) \\\ & \Rightarrow S=-\dfrac{1}{2}\left( \dfrac{{{3}^{n+1}}-3-2n{{.3}^{n+1}}}{2} \right) \\\ & \Rightarrow S=-\dfrac{1}{2}\left( \dfrac{{{3}^{n+1}}\left( 1-2n \right)-3}{2} \right) \\\ \end{aligned}$
Multiplying $-\dfrac{1}{2}$ inside the bracket in the above equation we get,
$S=\dfrac{{{3}^{n+1}}\left( 2n-1 \right)+3}{4}$
Hence, we get the summation of series $1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}$ as:
$S=\dfrac{{{3}^{n+1}}\left( 2n-1 \right)+3}{4}$…….. Eq. (4)
It is given that the summation of the series is equal to:
$\dfrac{\left( 2n-1 \right){{3}^{a}}+b}{4}$………….. Eq. (5)
Comparing the above summation with the summation that we have got after solving the series or (comparing eq. (4) with eq. (5)) we get,
$\begin{aligned} & a=n+1 \\\ & b=3 \\\ \end{aligned}$
Hence, we have got the value of $a=n+1;b=3$.

Note: There is a shortcut to find the solution of the above problem that we are going to discuss below.
The equation which is given in the question is:
$1.3+{{2.3}^{2}}+{{3.3}^{3}}+.....+n{{.3}^{n}}=\dfrac{\left( 2n-1 \right){{3}^{a}}+b}{4}$
As “n” can be a positive integer so starting from $n=1$, substitute different values of n and see what values of “a” & “b” are we getting.
Substituting $n=1$ in the given equation we get,
$\begin{aligned} & 1.3=\dfrac{{{3}^{a}}+b}{4} \\\ & \Rightarrow 12={{3}^{a}}+b \\\ \end{aligned}$
By hit and trial, we can find the value of “a” and “b” as when $a=2;b=3$ then we get 12.
Here, we have taken $n=1$ so the value of “a” is equal to n + 1.
Substituting $n=2$ in the given equation we get,
$\begin{aligned} & 1.3+{{2.3}^{2}}=\dfrac{{{3.3}^{a}}+b}{4} \\\ & \Rightarrow 3+18=\dfrac{{{3.3}^{a}}+b}{4} \\\ \end{aligned}$
$\Rightarrow 21=\dfrac{{{3.3}^{a}}+b}{4}$
On cross-multiplying the above equation we get,
$84={{3.3}^{a}}+b$
Now, L.H.S is equal to R.H.S when $a=3;b=3$.
Here, we have taken $n=2$ so the value of “a” is equal to n + 1.
As you can see from the above substitutions for $n=1\And 2$, we have found that:
$\begin{aligned} & a=n+1 \\\ & b=3 \\\ \end{aligned}$