Question

Find the value of $ \alpha $ if $ \left| \alpha \right|<1 $ and $ y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+......\infty $ .
(a) $ y+\dfrac{1}{y} $
(b) $ \dfrac{y}{1-y} $
(c) $ y-\dfrac{1}{y} $
(d) $ \dfrac{y}{1+y} $

Answer 1

Hint: Group the terms containing odd powers of $ \alpha $ together and the terms containing even powers of $ \alpha $ together. Now, apply the formula for the sum of infinite terms of a G.P given as $ {{S}_{\infty }}=\dfrac{a}{1-r} $ , where $ {{S}_{\infty }} $ is the sum of infinite terms, a is the first term and r is the common ratio of the G.P. Find this sum for the two separate groups and take their difference to get the answer.

Complete step by step answer:
Here, we have been provided with the expression $ y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+......\infty $ and we have to find the value of $ \alpha $ in terms of y. Now, we can write the given expression as
 $ \Rightarrow y=\alpha -{{\alpha }^{2}}+{{\alpha }^{3}}-{{\alpha }^{4}}+......\infty $
Separating and grouping the terms of odd powers of $ \alpha $ together and the terms containing even powers of $ \alpha $ together, we get
 $ \Rightarrow y=\left( \alpha +{{\alpha }^{3}}+{{\alpha }^{5}}+....\infty \right)-\left( {{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+....\infty \right) $
Let us write the above expression as
 $ \Rightarrow y={{S}_{1}}-{{S}_{2}} $
Here, we have
\[\begin{align}
  & {{S}_{1}}=\alpha +{{\alpha }^{3}}+{{\alpha }^{5}}+....\infty \ldots \ldots \ldots \left( i \right) \\
 & {{S}_{2}}={{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+....\infty \ldots \ldots \ldots \left( ii \right) \\
\end{align}\]
So let us find the sums separately. From equation (i), we have
 $ {{S}_{1}}=\alpha +{{\alpha }^{3}}+{{\alpha }^{5}}+....\infty $
Clearly, we can see that the terms in the above expression are in G.P. whose first term is $ \alpha $ and the common ratio is $ \dfrac{{{\alpha }^{3}}}{\alpha }={{\alpha }^{2}} $ . Let us represent the first term with a and common ratio with r. Therefore, applying the formula of sum of infinite terms of a G.P, we have
 $ \Rightarrow {{S}_{\infty }}=\dfrac{a}{1-r} $ , where $ {{S}_{\infty }} $ is the sum of infinite terms.
Therefore, substituting the values of a and r, we get
\[\begin{align}
  & \Rightarrow {{S}_{\infty }}={{S}_{1}}=\dfrac{\alpha }{1-{{\alpha }^{2}}} \\
 & \Rightarrow {{S}_{1}}=\dfrac{\alpha }{1-{{\alpha }^{2}}}\ldots \ldots \ldots \left( iii \right) \\
\end{align}\]
Now, from equation (ii), we have,
 $ {{S}_{2}}={{\alpha }^{2}}+{{\alpha }^{4}}+{{\alpha }^{6}}+....\infty $
Clearly, we can see that the terms in the above expression are also in G.P. whose first term is $ {{\alpha }^{2}} $ and the common ratio is $ \dfrac{{{\alpha }^{4}}}{{{\alpha }^{2}}}={{\alpha }^{2}} $ . So, here also applying the formula for sum of infinite terms of a G.P, we have
\[\Rightarrow {{S}_{2}}=\dfrac{{{\alpha }^{2}}}{1-{{\alpha }^{2}}}\ldots \ldots \ldots \left( iv \right)\]
Subtracting equation (iv) from equation (iii), we get
\[\begin{align}
  & \Rightarrow {{S}_{1}}-{{S}_{2}}=\dfrac{\alpha }{1-{{\alpha }^{2}}}-\dfrac{{{\alpha }^{2}}}{1-{{\alpha }^{2}}} \\
 & \Rightarrow y=\dfrac{\alpha -{{\alpha }^{2}}}{1-{{\alpha }^{2}}} \\
 & \Rightarrow y=\dfrac{\alpha \left( 1-\alpha \right)}{\left( 1-\alpha \right)\left( 1+\alpha \right)} \\
\end{align}\]
Cancelling the common factors, we get
\[\Rightarrow y=\dfrac{\alpha }{\left( 1+\alpha \right)}\]
By cross multiplying, we get
 $ \begin{align}
  & \Rightarrow y+\alpha y=\alpha \\
 & \Rightarrow y=\alpha -\alpha y \\
 & \Rightarrow y=\alpha \left( 1-y \right) \\
 & \Rightarrow \alpha =\dfrac{y}{1-y} \\
\end{align} $
Hence, option (b) is the correct answer.

Note:
One may note that we do not need to separate and group the odd powers and even powers of $ \alpha $ in the expression of y. We can directly apply the formula for the sum of infinite terms of the G.P. by considering the first term (a) as $ \alpha $ and the common ratio (r) as $ -\alpha $ . But here we have separated the terms because at many places you will need the above-used separation method which should be known. Remember that here we have to find the value of $ \alpha $ in terms of y and not y in terms of $ \alpha $ , so do not make a mistake here. At last, note that here we were provided with the condition $ \left| \alpha \right|<1 $ and that is why we applied the formula $ {{S}_{\infty }}=\dfrac{a}{1-r} $ , if we would have been provided with $ \left| \alpha \right|>1 $ then we would have applied $ {{S}_{\infty }}=\dfrac{a}{r-1} $ .