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Question: The value of \({{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7...

The value of 37C0+47C1+57C2+..........+107C7{{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7}} is__________
A) 10(2)610{{\left( 2 \right)}^{6}}
B) 13(2)713{{\left( 2 \right)}^{7}}
C) 14(2)614{{\left( 2 \right)}^{6}}
D) 13(2)613{{\left( 2 \right)}^{6}}

Explanation

Solution

Here in this question we have been asked to find the value of the given summation 37C0+47C1+57C2+..........+107C7{{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7}} in a simplified form. For answering this question we will use r.nCr=nn1Crr{{.}^{n}}{{C}_{r}}={{n}^{n-1}}{{C}_{r}} .

Complete step by step answer:
Now considering from the question we have been asked to find the value of the given summation 37C0+47C1+57C2+..........+107C7{{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7}} in a simplified form.
For answering this question firstly we will observe the pattern involved.
By observing the pattern we can simply write the rth{{r}^{th}} term of the given pattern will be tr=(r+3)7Cr{{t}_{r}}={{\left( r+3 \right)}^{7}}{{C}_{r}} .
Hence we can write it simply as 37C0+47C1+57C2+..........+107C7=r=07(r+3)7Cr{{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7}}=\sum\limits_{r=0}^{7}{{{\left( r+3 \right)}^{7}}{{C}_{r}}} .
By further simplifying the above expression we will have r=07r7Cr+3r=077Cr\Rightarrow \sum\limits_{r=0}^{7}{{{r}^{7}}{{C}_{r}}}+3\sum\limits_{r=0}^{7}{^{7}{{C}_{r}}} ,
From the basic concepts of combinations we know that r.nCr=nn1Crr{{.}^{n}}{{C}_{r}}={{n}^{n-1}}{{C}_{r}} now we will use this to simplify the expression further.
By using the above expression we will have 7r=066Cr+3r=077Cr\Rightarrow 7\sum\limits_{r=0}^{6}{^{6}{{C}_{r}}}+3\sum\limits_{r=0}^{7}{^{7}{{C}_{r}}} .
From the basic concepts of combinations we know that r=0nnCr=2n\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}={{2}^{n}} now we will use this formula to simplify the above expression further.
Now by further simplifying the above expression using the formula we will have 7(26)+3(27)\Rightarrow 7\left( {{2}^{6}} \right)+3\left( {{2}^{7}} \right) .
Now we will take out common 26{{2}^{6}} from the expression after doing that we will have 26(7+3(2))\Rightarrow {{2}^{6}}\left( 7+3\left( 2 \right) \right) .
Now by further simplifying this we will have
26(7+6) 13(26) \begin{aligned} & \Rightarrow {{2}^{6}}\left( 7+6 \right) \\\ & \Rightarrow 13\left( {{2}^{6}} \right) \\\ \end{aligned} .
Therefore we can conclude that the value of 37C0+47C1+57C2+..........+107C7{{3}^{7}}{{C}_{0}}+{{4}^{7}}{{C}_{1}}+{{5}^{7}}{{C}_{2}}+..........+{{10}^{7}}{{C}_{7}}\, is 13(2)613{{\left( 2 \right)}^{6}} .

So, the correct answer is “Option D”.

Note: During the process of answering questions of this type we should be sure with the concepts that we are applying and calculations that we are going to perform in between the steps. Now we can generalize this as r=0n(r+x)nCr=2n1(n+2x)\sum\limits_{r=0}^{n}{{{\left( r+x \right)}^{n}}{{C}_{r}}}={{2}^{n-1}}\left( n+2x \right) .