If ${{C}_{K}}$ is the coefficient of ${{x}^{K}}$ in ${{\left( 1+x \right)}^{2005}}$ and if $a,d\in R$ then $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}=}$
A. $\left( 2a+2005d \right){{2}^{2004}}$
B. $\left( 2a+2005d \right){{2}^{2005}}$
C. $\left( 2a+2004d \right){{2}^{2005}}$
D. $\left( 2a+2004d \right){{2}^{2004}}$
Answer
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Hint: We here need to find the value of $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}$. For this we have been given that a and d are two real numbers and ${{C}_{K}}$ is the coefficient of ${{x}^{K}}$ in ${{\left( 1+x \right)}^{2005}}$. Since, ${{C}_{K}}$ is the coefficient of ${{x}^{K}}$ in ${{\left( 1+x \right)}^{2005}}$, we will get the value of ${{C}_{K}}$ as:
${{C}_{K}}=\dfrac{2005!}{K!\left( 2005-K \right)!}$
Then, we will break the given sigma into two sigmas $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$ and $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$ respectively by the property $\sum{\left( x+y \right).Z}=\sum{\left( xZ \right)+\sum{\left( yZ \right)}}$. Then we will solve both these sigmas separately. We will take out the constants from the sigma and try to make what’s left over inside in the form of $\sum\limits_{r=0}^{n}{{{C}_{r}}}$ as this value is given as: $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$. Hence, we will obtain the values of both these separate sigmas. Adding those values, we will get our desired result.
Complete step by step answer:
Now, we have been given that ${{C}_{K}}$ is the coefficient of ${{x}^{K}}$ in ${{\left( 1+x \right)}^{2005}}$.
Now, we know that the coefficient of ${{x}^{r}}$ in ${{\left( 1+x \right)}^{n}}$ is given as:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Hence, the value of ${{C}_{K}}$ is given as:
${{C}_{K}}=\dfrac{2005!}{K!\left( 2005-K \right)!}$
Now, we have been asked to find the value of $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}$ where a and d are two real numbers.
Now, we know that $\sum{\left( x+y \right).Z}=\sum{\left( xZ \right)+\sum{\left( yZ \right)}}$
Thus, applying this property in $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}$, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}} \\
& \Rightarrow \sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}+\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
\end{align}$
Thus, we have two sigmas to solve which when combined will give us our required answer.
We will now first solve the sigma $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$.
Now, since ‘a’ is a real constant, we can take it outside the sigma by the property $\sum{\left( kx \right)}=k\sum{x}$ where ‘k’ is a constant and x is a variable.
Hence, we get the value of $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$ as:
$\begin{align}
& \sum\limits_{K=0}^{2005}{(a.{{C}_{K}})} \\
& \Rightarrow a\sum\limits_{K=0}^{2005}{{{C}_{K}}} \\
\end{align}$
Now, we know that the value of $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$
Here, r=K and n=2005.
Thus, we get:
$\begin{align}
& a\sum\limits_{K=0}^{2005}{{{C}_{K}}} \\
& \Rightarrow a\left( {{2}^{2005}} \right) \\
& \Rightarrow a{{.2}^{2005}} \\
\end{align}$
Hence, the value of the first sigma is equal to $a{{.2}^{2005}}$
Now, we will solve our second sigma $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$.
Here, out of K and d, d is a real constant. Thus, we can take d out of the sigma by the property $\sum{\left( kx \right)}=k\sum{x}$ where ‘k’ is a constant and x is a variable.
Thus, we get the value of $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$ as:
$\begin{align}
& \sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K{{C}_{K}}} \\
\end{align}$
Now, we will expand the value of ${{C}_{K}}$ as mentioned above:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{K{{C}_{K}}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K.\dfrac{2005!}{K!\left( 2005-K \right)!}} \\
\end{align}$
Now, we know that n! can be also expressed as:
$n\left( n-1 \right)!=n\left( n-1 \right)\left( n-2 \right)!=...$
Thus, expanding the factorials in the above expansion of the sigma as mentioned here we get:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{K.\dfrac{2005!}{K!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K.\dfrac{2005.\left( 2005-1 \right)!}{K\left( K-1 \right)!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2005-K \right)!}} \\
\end{align}$
Now, we can write 2005-K as:
2004-(K-1)
Hence, writing 2005-K in the way mentioned we get:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
& \Rightarrow d.2005\sum\limits_{K=0}^{2005}{\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
\end{align}$
Now, we can see that $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ is in the form of $\dfrac{n!}{r!\left( n-r \right)!}$ which is equal to $^{n}{{C}_{r}}$ .
Thus, $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ will be equal to:
$^{2004}{{C}_{K-1}}$
Hence, putting the value of $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ in the required value of sigma we get:
$\begin{align}
& d.2005\sum\limits_{K=0}^{2005}{\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
& \Rightarrow 2005d\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}} \\
\end{align}$
Now, we know that $\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}}$ will be equal to ${{2}^{2004}}$ by the property $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$.
Hence, we get our sigma as:
$\begin{align}
& 2005d\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}} \\
& \Rightarrow 2005d\left( {{2}^{2004}} \right) \\
& \Rightarrow {{2}^{2004}}.2005d \\
\end{align}$
Hence, we get the value of our second sigma as ${{2}^{2004}}.2005d$.
Now, by adding the values of both these sigmas obtained, we get the value of our required sigma.
Hence, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}=\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}+\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
& \Rightarrow \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}={{2}^{2005}}.a+{{2}^{2004}}.2005d \\
\end{align}$
By taking ${{2}^{2004}}$ common, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}={{2}^{2005}}.a+{{2}^{2004}}.2005d \\
& \Rightarrow \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}=\left( 2a+2005d \right){{2}^{2004}} \\
\end{align}$
Hence, the required answer is $\left( 2a+2005d \right){{2}^{2004}}$
So, the correct answer is “Option A”.
Note: We have here used the property $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$.
Its proof is given as follows:
We know that the binomial expansion ${{\left( x+y \right)}^{n}}$ in sigma form is expressed as:
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}}$
If we keep both x=1 and y=1, we will get:
$\begin{align}
& {{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}} \\
& \Rightarrow {{\left( 1+1 \right)}^{2}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{.1}^{n-r}}{{.1}^{r}}} \\
& \Rightarrow \sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}={{2}^{n}} \\
\end{align}$
Hence, proved.
${{C}_{K}}=\dfrac{2005!}{K!\left( 2005-K \right)!}$
Then, we will break the given sigma into two sigmas $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$ and $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$ respectively by the property $\sum{\left( x+y \right).Z}=\sum{\left( xZ \right)+\sum{\left( yZ \right)}}$. Then we will solve both these sigmas separately. We will take out the constants from the sigma and try to make what’s left over inside in the form of $\sum\limits_{r=0}^{n}{{{C}_{r}}}$ as this value is given as: $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$. Hence, we will obtain the values of both these separate sigmas. Adding those values, we will get our desired result.
Complete step by step answer:
Now, we have been given that ${{C}_{K}}$ is the coefficient of ${{x}^{K}}$ in ${{\left( 1+x \right)}^{2005}}$.
Now, we know that the coefficient of ${{x}^{r}}$ in ${{\left( 1+x \right)}^{n}}$ is given as:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Hence, the value of ${{C}_{K}}$ is given as:
${{C}_{K}}=\dfrac{2005!}{K!\left( 2005-K \right)!}$
Now, we have been asked to find the value of $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}$ where a and d are two real numbers.
Now, we know that $\sum{\left( x+y \right).Z}=\sum{\left( xZ \right)+\sum{\left( yZ \right)}}$
Thus, applying this property in $\sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}$, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}} \\
& \Rightarrow \sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}+\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
\end{align}$
Thus, we have two sigmas to solve which when combined will give us our required answer.
We will now first solve the sigma $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$.
Now, since ‘a’ is a real constant, we can take it outside the sigma by the property $\sum{\left( kx \right)}=k\sum{x}$ where ‘k’ is a constant and x is a variable.
Hence, we get the value of $\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}$ as:
$\begin{align}
& \sum\limits_{K=0}^{2005}{(a.{{C}_{K}})} \\
& \Rightarrow a\sum\limits_{K=0}^{2005}{{{C}_{K}}} \\
\end{align}$
Now, we know that the value of $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$
Here, r=K and n=2005.
Thus, we get:
$\begin{align}
& a\sum\limits_{K=0}^{2005}{{{C}_{K}}} \\
& \Rightarrow a\left( {{2}^{2005}} \right) \\
& \Rightarrow a{{.2}^{2005}} \\
\end{align}$
Hence, the value of the first sigma is equal to $a{{.2}^{2005}}$
Now, we will solve our second sigma $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$.
Here, out of K and d, d is a real constant. Thus, we can take d out of the sigma by the property $\sum{\left( kx \right)}=k\sum{x}$ where ‘k’ is a constant and x is a variable.
Thus, we get the value of $\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})}$ as:
$\begin{align}
& \sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K{{C}_{K}}} \\
\end{align}$
Now, we will expand the value of ${{C}_{K}}$ as mentioned above:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{K{{C}_{K}}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K.\dfrac{2005!}{K!\left( 2005-K \right)!}} \\
\end{align}$
Now, we know that n! can be also expressed as:
$n\left( n-1 \right)!=n\left( n-1 \right)\left( n-2 \right)!=...$
Thus, expanding the factorials in the above expansion of the sigma as mentioned here we get:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{K.\dfrac{2005!}{K!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{K.\dfrac{2005.\left( 2005-1 \right)!}{K\left( K-1 \right)!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2005-K \right)!}} \\
\end{align}$
Now, we can write 2005-K as:
2004-(K-1)
Hence, writing 2005-K in the way mentioned we get:
$\begin{align}
& d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2005-K \right)!}} \\
& \Rightarrow d\sum\limits_{K=0}^{2005}{\dfrac{2005.2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
& \Rightarrow d.2005\sum\limits_{K=0}^{2005}{\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
\end{align}$
Now, we can see that $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ is in the form of $\dfrac{n!}{r!\left( n-r \right)!}$ which is equal to $^{n}{{C}_{r}}$ .
Thus, $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ will be equal to:
$^{2004}{{C}_{K-1}}$
Hence, putting the value of $\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}$ in the required value of sigma we get:
$\begin{align}
& d.2005\sum\limits_{K=0}^{2005}{\dfrac{2004!}{\left( K-1 \right)!\left( 2004-\left( K-1 \right) \right)!}} \\
& \Rightarrow 2005d\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}} \\
\end{align}$
Now, we know that $\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}}$ will be equal to ${{2}^{2004}}$ by the property $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$.
Hence, we get our sigma as:
$\begin{align}
& 2005d\sum\limits_{K=0}^{2005}{^{2004}{{C}_{K-1}}} \\
& \Rightarrow 2005d\left( {{2}^{2004}} \right) \\
& \Rightarrow {{2}^{2004}}.2005d \\
\end{align}$
Hence, we get the value of our second sigma as ${{2}^{2004}}.2005d$.
Now, by adding the values of both these sigmas obtained, we get the value of our required sigma.
Hence, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}=\sum\limits_{K=0}^{2005}{(a.{{C}_{K}})}+\sum\limits_{K=0}^{2005}{(Kd.{{C}_{K}})} \\
& \Rightarrow \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}={{2}^{2005}}.a+{{2}^{2004}}.2005d \\
\end{align}$
By taking ${{2}^{2004}}$ common, we get:
$\begin{align}
& \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}={{2}^{2005}}.a+{{2}^{2004}}.2005d \\
& \Rightarrow \sum\limits_{K=0}^{2005}{\left( a+Kd \right).{{C}_{K}}}=\left( 2a+2005d \right){{2}^{2004}} \\
\end{align}$
Hence, the required answer is $\left( 2a+2005d \right){{2}^{2004}}$
So, the correct answer is “Option A”.
Note: We have here used the property $\sum\limits_{r=0}^{n}{{{C}_{r}}}={{2}^{n}}$.
Its proof is given as follows:
We know that the binomial expansion ${{\left( x+y \right)}^{n}}$ in sigma form is expressed as:
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}}$
If we keep both x=1 and y=1, we will get:
$\begin{align}
& {{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}} \\
& \Rightarrow {{\left( 1+1 \right)}^{2}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{.1}^{n-r}}{{.1}^{r}}} \\
& \Rightarrow \sum\limits_{r=0}^{n}{^{n}{{C}_{r}}}={{2}^{n}} \\
\end{align}$
Hence, proved.
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