Question

The value of $\sum\limits_{0\ \le \ i\ \ \le \ j\ \le \ 10}{\sum{\left( ^{10}{{\text{C}}_{j}} \right)\left( ^{j}{{\text{C}}_{j}} \right)}}$ is equal to
A. ${{3}^{10}}$
B. ${{3}^{10}}-1$
C. ${{2}^{10}}$
D. ${{2}^{10}}-1$

Answer 1

Hint: First expand the summation with all possible values then use the idea of identity that,
$^{n}{{\text{C}}_{\text{0}}}{{+}^{n}}{{\text{C}}_{1}}{{+}^{n}}{{\text{C}}_{2}}+........{{+}^{n}}{{\text{C}}_{n}}\ =\ {{2}^{n}}$ then try to use the theorem of Binomial expression,
$^{n}{{\text{C}}_{\text{0}}}{{x}^{0}}{{+}^{n}}{{\text{C}}_{1}}{{x}^{1}}{{+}^{n}}{{\text{C}}_{2}}{{x}^{2}}+........{{+}^{n}}{{\text{C}}_{n}}{{x}^{n}}$ to finally get the desired results.

Complete step by step answer:

In the question we have been asked to find $\sum\limits_{0\ \le \ i\ \ \le \ j\ \le \ 10}{\sum{\left( ^{10}{{\text{C}}_{j}} \right)\left( ^{j}{{\text{C}}_{j}} \right)}}$
Now, as we know that the expression is,
$\sum\limits_{0\ \le \ i\ \ \le \ j\ \le \ 10}{\sum{\left( ^{10}{{\text{C}}_{j}} \right)\left( ^{j}{{\text{C}}_{j}} \right)}}$
So, we will break it and we can write it as,
^{10}{{\text{C}}_{\text{0}}}{{+}^{10}}{{\text{C}}_{1}}\left( ^{1}{{\text{C}}_{\text{0}}}{{+}^{1}}{{\text{C}}_{1}} \right){{+}^{10}}{{\text{C}}_{2}}\left( ^{2}{{\text{C}}_{0}}{{+}^{2}}{{\text{C}}_{1}}{{+}^{2}}{{\text{C}}_{2}} \right){{+}^{10}}{{\text{C}}_{3}}\left( ^{3}{{\text{C}}_{0}}{{+}^{3}}{{\text{C}}_{1}}{{+}^{3}}{{\text{C}}_{2}}{{+}^{3}}{{\text{C}}_{3}} \right)+...$$$${{+}^{10}}{{\text{C}}_{4}}\left( ^{4}{{\text{C}}_{0}}{{+}^{4}}{{\text{C}}_{1}}{{+}^{4}}{{\text{C}}_{2}}{{+}^{4}}{{\text{C}}_{3}}{{+}^{4}}{{\text{C}}_{4}} \right)+.......{{+}^{10}}{{\text{C}}_{10}}\left( ^{10}{{\text{C}}_{0}}{{+}^{10}}{{\text{C}}_{1}}{{+}^{10}}{{\text{C}}_{2}}+{{..........}^{10}}{{\text{C}}_{10}} \right)
So, we can rewrite the above expression using formula,
$^{n}{{\text{C}}_{\text{0}}}{{+}^{n}}{{\text{C}}_{1}}{{+}^{n}}{{\text{C}}_{2}}+........{{+}^{n}}{{\text{C}}_{n}}\ =\ {{2}^{n}}$
So, $^{1}{{\text{C}}_{\text{0}}}{{+}^{1}}{{\text{C}}_{1}}=\ {{2}^{1}}$
$^{2}{{\text{C}}_{\text{0}}}{{+}^{2}}{{\text{C}}_{1}}{{+}^{2}}{{\text{C}}_{2}}=\ 2$
$^{3}{{\text{C}}_{\text{0}}}{{+}^{3}}{{\text{C}}_{1}}{{+}^{3}}{{\text{C}}_{2}}{{+}^{3}}{{\text{C}}_{3}}=\ {{2}^{3}}$
$^{4}{{\text{C}}_{\text{0}}}{{+}^{4}}{{\text{C}}_{1}}{{+}^{4}}{{\text{C}}_{2}}{{+}^{4}}{{\text{C}}_{3}}{{+}^{4}}{{\text{C}}_{4}}=\ {{2}^{4}}$
Just like this we can also represent,
$^{10}{{\text{C}}_{0}}{{+}^{10}}{{\text{C}}_{1}}+{{..........}^{10}}{{\text{C}}_{10}}=\ {{2}^{10}}$
Hence, we can write expression as,
$^{10}{{\text{C}}_{0}}\centerdot {{2}^{0}}{{+}^{10}}{{\text{C}}_{1}}\centerdot {{2}^{1}}{{+}^{10}}{{\text{C}}_{2}}\centerdot {{2}^{2}}{{+}^{10}}{{\text{C}}_{3}}\centerdot {{2}^{3}}+..........{{+}^{10}}{{\text{C}}_{10}}\centerdot {{2}^{10}}$.
So, we can write it as,
${{\left( 1+2 \right)}^{10}}\ =\ {{3}^{10}}$
Or, ${{\left( 1+2 \right)}^{10}}\ ={{\ }^{10}}{{\text{C}}_{0}}\centerdot {{2}^{0}}{{+}^{10}}{{\text{C}}_{1}}\centerdot {{2}^{1}}{{+}^{10}}{{\text{C}}_{2}}\centerdot {{2}^{2}}{{+}^{10}}{{\text{C}}_{3}}\centerdot {{2}^{3}}+..........{{+}^{10}}{{\text{C}}_{10}}\centerdot {{2}^{10}}$ .
Hence, the answer is ${{\left( 1+2 \right)}^{10}}$ which is ${{3}^{10}}$.
Hence, the correct option is ‘A’.

Note: Students should have an idea of how to reconvert back from expansion of term to back into a single term with power. They should also be careful about calculation too.