Question

If x + y = 1, then $\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$ equals: -
(a) $nxy$
(b) $nx\left( x+yn \right)$
(c) $n\left( nx+y \right)$
(d) 1
(e) $nx\left( nx+y \right)$

Answer 1

Write the expanded form of ${{\left( y+x \right)}^{n}}$ in term of ${}^{n}{{C}_{r}}$. Convert this expanded form into general form by using the summation sign ($\sum{{}}$). Now, partially differentiate both sides of the relation with respect to x, keeping y constant, and after that multiply both sides with x. Now, again partially differentiate both the sides of the obtained expression with respect to x, keeping y constant, and multiply both sides with x. Substitute the given value x + y = 1 to get the correct answer.

Complete answer:
Here, we have been provided with the relation x + y = 1 and we have to find the value of the expression: - $\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$.
Now, we know that the expanded form of ${{\left( x+y \right)}^{n}}$ is given as: -
$\Rightarrow {{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+.......+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}$
In general form we can write the above expression using summation sign as: -
$\Rightarrow {{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$ - (1)
But we can see that in the question we have to find the value of the expression in which the exponent of y is (n – r) and x is r, which is just the reverse of the expression we have obtained in equation (1). So, on interchanging the places of x and y in the L.H.S, we get,
$\Rightarrow {{\left( y+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$
Now, partially differentiating the above relation with respect to x, keeping y constant, we get,

& \Rightarrow n{{\left( y+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}.r{{x}^{r-1}}{{y}^{n-r}}} \\\ & \Rightarrow n{{\left( y+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r-1}}{{y}^{n-r}}} \\\ \end{aligned}$$ Multiplying both sides with x, we get, $$\Rightarrow nx{{\left( y+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$$ Again partially differentiating the above relation with respect to x, keeping y constant, we get, $$\Rightarrow n\dfrac{d\left[ x{{\left( y+x \right)}^{n-1}} \right]}{dx}=\sum\limits_{r=0}^{n}{r.{{C}_{r}}\times r{{x}^{r-1}}{{y}^{n-r}}}$$ Using the product rule of differentiation in the L.H.S given as: - $$\dfrac{d\left( u\times v \right)}{dx}=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}$$, we get, $$\Rightarrow n\times \left[ x\left( n-1 \right){{\left( y+x \right)}^{n-2}}+{{\left( y+x \right)}^{n-1}} \right]=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r-1}}{{y}^{n-r}}}$$ Multiplying both sides with x, we get, $$\Rightarrow nx\left[ x\left( n-1 \right){{\left( y+x \right)}^{n-2}}+{{\left( y+x \right)}^{n-1}} \right]=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r-1}}{{y}^{n-r}}\times x}$$ Substituting x + y = 1, we get, $$\begin{aligned} & \Rightarrow nx\left[ x\left( n-1 \right)\times 1+1 \right]=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}} \\\ & \Rightarrow nx\left[ nx-x+1 \right]=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}} \\\ \end{aligned}$$ Now, we have, x + y = 1, $$\begin{aligned} & \Rightarrow y=1-x \\\ & \Rightarrow y=-x+1 \\\ \end{aligned}$$ So, the obtained relation becomes, $$\Rightarrow nx\left[ nx+y \right]=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}{{y}^{n-r}}}$$ **Hence, option (e) is the correct answer.** **Note:** One may note that the value of both the expressions $${{\left( x+y \right)}^{n}}$$ and $${{\left( y+x \right)}^{n}}$$ are the same and their expanded as well as general forms are also the same. This is due to the property of combination given as: - $${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$$. So, do not get confused by this. We have used this conversion because it was necessary to get an expression provided in the question. Now, you may note that we have differentiated the relations with respect to x. This was done to get $${{r}^{2}}$$. Here, we had considered y as a constant to differentiate the function with respect to x. This process is known as partial differentiation.