
Let we are given the series as \[a = 111...1\]\[(55\] digits\[)\], \[b = 1 + 10 + {10^2} + ... + {10^4}\], \[c = 1 + {10^5} + {10^{10}} + {10^{15}} + ... + {10^{50}}\], then
a). \[a = b + c\]
b). \[a = bc\]
c). \[b = ac\]
d). \[c = ab\]
Answer
488.7k+ views
Hint: First we take intuition from smaller sets of numbers about the type of representation they have and then try to do it for bigger numbers. We also use G.P. series for summing the given numbers and drawing a relationship between them.
Complete step-by-step solution:
We have \[a = 111...1\]\[(55\] digits \[)\], \[b = 1 + 10 + {10^2} + ... + {10^4}\], \[c = 1 + {10^5} + {10^{10}} + {10^{15}} + ... + {10^{50}}\],
Let us try to represent a smaller repeated number with \[1\] as its repeated digit.
\[111 = 1 + 10 + {10^2}\].
Now let us try to do the same for \[a = 111...1\]\[(55\] digits\[)\]i.e.
\[a = 1 + 10 + {10^2} + ... + {10^{54}}\]
Now using G.P. series for the summation of \[a = 1 + 10 + {10^2} + ... + {10^{54}}\] we have,\[A = 1\],\[n = 55\] and \[r = 10\] then,
\[a = \dfrac{{A.{r^n} - 1}}{{r - 1}}\]
So, \[a = \dfrac{{{{10}^{55}} - 1}}{{10 - 1}}\]
Now let us sum the other numbers using G.P. property for \[b\] we have,\[A = 1\], \[n = 5\] and \[r = 10\] then,
\[b = \dfrac{{{{10}^5} - 1}}{{10 - 1}}\]
Again, for \[c\] using G.P. property we have \[A = 1\], \[n = 11\] and \[r = {10^5}\] then,
\[
c = \dfrac{{{{({{10}^5})}^{11}} - 1}}{{{{10}^5} - 1}} \\
c = \dfrac{{{{10}^{55}} - 1}}{{{{10}^5} - 1}} \]
Now using a bit of observation and analysis we notice that multiplying \[b,c\] we get,
\[
= \dfrac{{{{10}^5} - 1}}{{10 - 1}}.\dfrac{{{{10}^{55}} - 1}}{{{{10}^5} - 1}} \\
= \dfrac{{{{10}^{55}} - 1}}{{10 - 1}} \]
That is equal to \[a = \dfrac{{{{10}^{55}} - 1}}{{10 - 1}}\].
Thus from the above observation $a=bc$.
Hence the correct option is (b).
Additional information: Geometric series performed a critical function in the early development of calculus, and retain as a vital part of the observance of the convergence of series. Geometric series are used during arithmetic. They have important applications in physics, engineering, biology, economics, computer science, queuing concepts, and finance. Geometric series are one of the handiest examples of infinite series with finite sums, despite the fact that not all of them have these assets. Population growth and nuclear decay are some real life examples of Geometric progression where their growth or decay are seen to be following the trend and pattern of Geometric series.
Note: It is very important that we have the basic knowledge of series and sequence and know their pattern and basic properties of summation, common difference etc. It is also necessary that moving forward to gather the idea of intuition where we could probably get an idea of what can be the probable solution to a problem.
Complete step-by-step solution:
We have \[a = 111...1\]\[(55\] digits \[)\], \[b = 1 + 10 + {10^2} + ... + {10^4}\], \[c = 1 + {10^5} + {10^{10}} + {10^{15}} + ... + {10^{50}}\],
Let us try to represent a smaller repeated number with \[1\] as its repeated digit.
\[111 = 1 + 10 + {10^2}\].
Now let us try to do the same for \[a = 111...1\]\[(55\] digits\[)\]i.e.
\[a = 1 + 10 + {10^2} + ... + {10^{54}}\]
Now using G.P. series for the summation of \[a = 1 + 10 + {10^2} + ... + {10^{54}}\] we have,\[A = 1\],\[n = 55\] and \[r = 10\] then,
\[a = \dfrac{{A.{r^n} - 1}}{{r - 1}}\]
So, \[a = \dfrac{{{{10}^{55}} - 1}}{{10 - 1}}\]
Now let us sum the other numbers using G.P. property for \[b\] we have,\[A = 1\], \[n = 5\] and \[r = 10\] then,
\[b = \dfrac{{{{10}^5} - 1}}{{10 - 1}}\]
Again, for \[c\] using G.P. property we have \[A = 1\], \[n = 11\] and \[r = {10^5}\] then,
\[
c = \dfrac{{{{({{10}^5})}^{11}} - 1}}{{{{10}^5} - 1}} \\
c = \dfrac{{{{10}^{55}} - 1}}{{{{10}^5} - 1}} \]
Now using a bit of observation and analysis we notice that multiplying \[b,c\] we get,
\[
= \dfrac{{{{10}^5} - 1}}{{10 - 1}}.\dfrac{{{{10}^{55}} - 1}}{{{{10}^5} - 1}} \\
= \dfrac{{{{10}^{55}} - 1}}{{10 - 1}} \]
That is equal to \[a = \dfrac{{{{10}^{55}} - 1}}{{10 - 1}}\].
Thus from the above observation $a=bc$.
Hence the correct option is (b).
Additional information: Geometric series performed a critical function in the early development of calculus, and retain as a vital part of the observance of the convergence of series. Geometric series are used during arithmetic. They have important applications in physics, engineering, biology, economics, computer science, queuing concepts, and finance. Geometric series are one of the handiest examples of infinite series with finite sums, despite the fact that not all of them have these assets. Population growth and nuclear decay are some real life examples of Geometric progression where their growth or decay are seen to be following the trend and pattern of Geometric series.
Note: It is very important that we have the basic knowledge of series and sequence and know their pattern and basic properties of summation, common difference etc. It is also necessary that moving forward to gather the idea of intuition where we could probably get an idea of what can be the probable solution to a problem.
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