For r = 0, 1, 2,.….10, let \[{{A}_{r}},{{B}_{r}},{{\text{C}}_{r}}\] denote respectively the coefficient of \[{{x}^{r}}\] in the expansions of
\[{{\left( 1+x \right)}^{10}},{{\left( 1+x \right)}^{20}},{{\left( 1+x \right)}^{30}}\] . Then find the value of:
\[\sum\limits_{r=1}^{10}{{{A}_{r}}\left( {{B}_{r}}{{B}_{10}}-{{C}_{10}}{{A}_{r}} \right)}\]
(a) \[{{B}_{10}}-{{C}_{10}}\]
(b) \[{{A}_{10}}\left( B_{10}^{2}-{{C}_{10}}{{A}_{10}} \right)\]
(c) 0
(d) \[{{C}_{10}}-{{B}_{10}}\]
Answer
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Hint: First solve the term inside sigma algebraically, and then separate it into two terms.
After that see if you can convert the term sigma into coefficients from any expansion and urm everything in terms of A,B, C. \[\]
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]. By using this find the A, B, C and their relation.
Complete step-by-step answer:
By using binomial theorem:
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]
By using above theorem,
\[{{A}_{r}}={}^{10}{{C}_{r}}.....\left( 1 \right)\]
By applying this to find B, C, we get:
\[\begin{align}
& {{B}_{r}}={}^{20}{{C}_{r}}.....\left( 2 \right) \\
& {{C}_{r}}={}^{30}{{C}_{r}}.....\left( 3 \right) \\
\end{align}\]
We require the value of expression:
\[\sum\limits_{r=01}^{10}{{{A}_{r}}\left( {{B}_{r}}{{B}_{10}}-{{C}_{10}}{{A}_{r}} \right)}\]
By simplifying, we get:
\[\sum\limits_{r=1}^{10}{{{A}_{r}}{{B}_{r}}{{B}_{10}}-{{C}_{10}}A_{r}^{2}}\]
By separating it into 2 terms, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{{A}_{r}}{{B}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{A}_{r}}^{2}}\]
By substituting equation (1), equation (2), and equation (3) into the expression, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}.....\left( 4 \right)\].
Now we will separate terms from this expression.
Case-1
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}.....\left( 5 \right)\]
We know combinations property:
\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]
By using the above combinations property, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\]
As we can see, by binomial theorem we can say
\[{{\left( 1+x \right)}^{30}}={{\left( 1+x \right)}^{10}}{{\left( 1+x \right)}^{20}}\]
For finding the coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\] by above expression we will have 10 cases.
\[1.{{x}^{20}},x.{{x}^{19}},.......,{{x}^{10}}.{{x}^{10}}\]
By adding the coefficients of all cases above, we get:
\[\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\] \[\]
The value of \[\sum\limits_{r=0}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\] is equal to coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\].
As r = 1 is given instead of r = 0, we need to subtract 1.
So the equation (5) turns out to be:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}=\] \[{{B}_{10}}.\](Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\] - 1).
Case-2
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{r}}}.....\left( 6 \right)\]
We know combinations property:
\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]
By using the above combinations property, we get:
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\]
As we can see, by binomial theorem we can say
\[{{\left( 1+x \right)}^{20}}={{\left( 1+x \right)}^{10}}{{\left( 1+x \right)}^{10}}\]
For finding the coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] by above expression we will have 10 cases.
\[1.{{x}^{10}},x.{{x}^{9}},.......,{{x}^{10}}.1\]
By adding the coefficients of all cases above, we get:
\[\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\] \[\]
The value of \[\sum\limits_{r=0}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\] is equal to coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\].
As r = 1 is given instead of r = 0, we need to subtract 1.
So the equation (5) turns out to be:
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{r}}}=\] \[{{C}_{10}}.\](Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] - 1).
Substitute these back into equation (4):
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{{B}_{10}}.\](Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\]-1) - \[{{C}_{10}}.\](Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] - 1)…..(7)
By binomial theorem:
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]
Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\]= \[{}^{30}{{C}_{20}}\]
Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\]=\[{}^{20}{{C}_{10}}\]
\[\begin{align}
& {{B}_{10}}={}^{20}{{C}_{10}} \\
& {{C}_{10}}={}^{30}{{C}_{10}} \\
\end{align}\]
By substituting these into equation (7), we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{20}{{C}_{10}}\left( {}^{30}{{C}_{20}}-1 \right)-{}^{30}{{C}_{10}}\left( {}^{20}{{C}_{10}}-1 \right)\]
By simplifying, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{20}{{C}_{10}}.{}^{30}{{C}_{20}}-{}^{20}{{C}_{10}}-{}^{30}{{C}_{10}}.{}^{20}{{C}_{10}}+{}^{30}{{C}_{10}}\]
Observe:
\[{}^{30}{{C}_{20}}={}^{30}{{C}_{10}}\]
By cancelling common terms, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{30}{{C}_{10}}-{}^{20}{{C}_{10}}\]
\[\begin{align}
& {{B}_{10}}={}^{20}{{C}_{10}} \\
& {{C}_{10}}={}^{30}{{C}_{10}} \\
\end{align}\]
By substituting the above values, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{{C}_{10}}-{{B}_{10}}\]
Therefore option (d) is correct.
Note: Applying Binomial theorem is a crucial step in the solution. Be careful while applying, if you write any number wrong you may lead to the wrong answer.
The idea of substituting B, C and getting them back is also important, just be careful while substituting if you do wrong you get negative of the answer which is also present in options. So don’t be confused.
After that see if you can convert the term sigma into coefficients from any expansion and urm everything in terms of A,B, C. \[\]
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]. By using this find the A, B, C and their relation.
Complete step-by-step answer:
By using binomial theorem:
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]
By using above theorem,
\[{{A}_{r}}={}^{10}{{C}_{r}}.....\left( 1 \right)\]
By applying this to find B, C, we get:
\[\begin{align}
& {{B}_{r}}={}^{20}{{C}_{r}}.....\left( 2 \right) \\
& {{C}_{r}}={}^{30}{{C}_{r}}.....\left( 3 \right) \\
\end{align}\]
We require the value of expression:
\[\sum\limits_{r=01}^{10}{{{A}_{r}}\left( {{B}_{r}}{{B}_{10}}-{{C}_{10}}{{A}_{r}} \right)}\]
By simplifying, we get:
\[\sum\limits_{r=1}^{10}{{{A}_{r}}{{B}_{r}}{{B}_{10}}-{{C}_{10}}A_{r}^{2}}\]
By separating it into 2 terms, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{{A}_{r}}{{B}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{A}_{r}}^{2}}\]
By substituting equation (1), equation (2), and equation (3) into the expression, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}.....\left( 4 \right)\].
Now we will separate terms from this expression.
Case-1
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}.....\left( 5 \right)\]
We know combinations property:
\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]
By using the above combinations property, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\]
As we can see, by binomial theorem we can say
\[{{\left( 1+x \right)}^{30}}={{\left( 1+x \right)}^{10}}{{\left( 1+x \right)}^{20}}\]
For finding the coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\] by above expression we will have 10 cases.
\[1.{{x}^{20}},x.{{x}^{19}},.......,{{x}^{10}}.{{x}^{10}}\]
By adding the coefficients of all cases above, we get:
\[\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\] \[\]
The value of \[\sum\limits_{r=0}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{20-r}}}\] is equal to coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\].
As r = 1 is given instead of r = 0, we need to subtract 1.
So the equation (5) turns out to be:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}=\] \[{{B}_{10}}.\](Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\] - 1).
Case-2
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{r}}}.....\left( 6 \right)\]
We know combinations property:
\[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]
By using the above combinations property, we get:
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\]
As we can see, by binomial theorem we can say
\[{{\left( 1+x \right)}^{20}}={{\left( 1+x \right)}^{10}}{{\left( 1+x \right)}^{10}}\]
For finding the coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] by above expression we will have 10 cases.
\[1.{{x}^{10}},x.{{x}^{9}},.......,{{x}^{10}}.1\]
By adding the coefficients of all cases above, we get:
\[\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\] \[\]
The value of \[\sum\limits_{r=0}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{10-r}}}\] is equal to coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\].
As r = 1 is given instead of r = 0, we need to subtract 1.
So the equation (5) turns out to be:
\[{{C}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{10}{{C}_{r}}}=\] \[{{C}_{10}}.\](Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] - 1).
Substitute these back into equation (4):
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{{B}_{10}}.\](Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\]-1) - \[{{C}_{10}}.\](Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\] - 1)…..(7)
By binomial theorem:
Coefficient of \[{{x}^{r}}\] in expansion of \[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{r}}\]
Coefficient of \[{{x}^{20}}\] in \[{{\left( 1+x \right)}^{30}}\]= \[{}^{30}{{C}_{20}}\]
Coefficient of \[{{x}^{10}}\] in \[{{\left( 1+x \right)}^{20}}\]=\[{}^{20}{{C}_{10}}\]
\[\begin{align}
& {{B}_{10}}={}^{20}{{C}_{10}} \\
& {{C}_{10}}={}^{30}{{C}_{10}} \\
\end{align}\]
By substituting these into equation (7), we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{20}{{C}_{10}}\left( {}^{30}{{C}_{20}}-1 \right)-{}^{30}{{C}_{10}}\left( {}^{20}{{C}_{10}}-1 \right)\]
By simplifying, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{20}{{C}_{10}}.{}^{30}{{C}_{20}}-{}^{20}{{C}_{10}}-{}^{30}{{C}_{10}}.{}^{20}{{C}_{10}}+{}^{30}{{C}_{10}}\]
Observe:
\[{}^{30}{{C}_{20}}={}^{30}{{C}_{10}}\]
By cancelling common terms, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{}^{30}{{C}_{10}}-{}^{20}{{C}_{10}}\]
\[\begin{align}
& {{B}_{10}}={}^{20}{{C}_{10}} \\
& {{C}_{10}}={}^{30}{{C}_{10}} \\
\end{align}\]
By substituting the above values, we get:
\[{{B}_{10}}\sum\limits_{r=1}^{10}{{}^{10}{{C}_{r}}{}^{20}{{C}_{r}}}-{{C}_{10}}\sum\limits_{r=1}^{10}{{{\left( {}^{10}{{C}_{r}} \right)}^{2}}}=\]\[{{C}_{10}}-{{B}_{10}}\]
Therefore option (d) is correct.
Note: Applying Binomial theorem is a crucial step in the solution. Be careful while applying, if you write any number wrong you may lead to the wrong answer.
The idea of substituting B, C and getting them back is also important, just be careful while substituting if you do wrong you get negative of the answer which is also present in options. So don’t be confused.
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