# If p + q = 1 then show that $\sum\nolimits_{r=0}^{n}{{{r}^{2}}{{c}_{r}}{{p}^{r}}.{{q}^{n-r}}=npq+{{n}^{2}}{{p}^{2}}}$.

Last updated date: 18th Mar 2023

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Answer

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Hint: Use the formula ${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}$ and try differentiating it successively. Then multiply with the ‘x’ term and simplify it to get the desired result.

In the question we have to consider the following equation,

${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}..............\left( i \right)$

Now, let’s differentiate equation (i) with respect to ‘x’, we will get,

$n{{\left( 1+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r-1}}}$

Now multiply by ‘x’ on both sides, we will get,

$nx{{\left( 1+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r}}}..............\left( ii \right)$

Now, let’s differentiate equation (ii) with respect to ‘x’ and using product rule of differentiation, i.e., \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], we get

$n{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right)x{{\left( 1+x \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r-1}}}$

Now multiply by ‘x’ both sides, we will get,

$nx{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right){{x}^{2}}{{\left( 1+x \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}}.........\left( iii \right)$

Now, in the equation (iii) we will substitute ‘x’ by $\left( \dfrac{p}{q} \right)$, so we get,

$\begin{align}

& n\left( \dfrac{p}{q} \right){{\left( 1+\dfrac{p}{q} \right)}^{n-1}}+n\left( n-1 \right){{\left( \dfrac{p}{q} \right)}^{2}}{{\left( 1+\dfrac{p}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}} \\

& \Rightarrow n\left( \dfrac{p}{q} \right){{\left( \dfrac{p+q}{q} \right)}^{n-1}}+n\left( n-1 \right)\left( \dfrac{{{p}^{2}}}{{{q}^{2}}} \right){{\left( \dfrac{p+q}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}} \\

\end{align}$

In the question we were given that ‘p + q = 1’, so substituting this in above equation, we get

$n\left( \dfrac{p}{q} \right){{\left( \dfrac{1}{q} \right)}^{n-1}}+n\left( n-1 \right)\left( \dfrac{{{p}^{2}}}{{{q}^{2}}} \right){{\left( \dfrac{1}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}}$

Combining the like terms, we get

$\Rightarrow n\dfrac{p}{{{q}^{n}}}+\dfrac{n\left( n-1 \right){{p}^{2}}}{{{q}^{n}}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}}$

Now, multiplying ${{q}^{n}}$on both side of the above equation we get,

$np+n\left( n-1 \right){{p}^{2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}$

We can also write like this

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np+n\left( n-1 \right){{p}^{2}}$

Combining the like terms, we get

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np\left( 1+np-p \right)$

We were given that ‘p + q = 1’, so, we can replace $\left( 1-p \right)$ by $q$, the above equation can be written as,

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np\left( q+np \right)$

Opening the bracket,w e get

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=npq+{{n}^{2}}{{p}^{2}}$

Hence Proved

Note: In these type of questions, student generally go wrong while differentiating;

${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}$with respect to $x$.

Another approach of this problem is

\[\sum\nolimits_{r=0}^{n}{{{r}^{2}}{{c}_{r}}{{p}^{r}}.{{q}^{n-r}}}\]

And convert this to the formula, ${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}$

In this way we can prove LHS is equal to RHS.

In the question we have to consider the following equation,

${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}..............\left( i \right)$

Now, let’s differentiate equation (i) with respect to ‘x’, we will get,

$n{{\left( 1+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r-1}}}$

Now multiply by ‘x’ on both sides, we will get,

$nx{{\left( 1+x \right)}^{n-1}}=\sum\limits_{r=0}^{n}{r{}^{n}{{C}_{r}}{{x}^{r}}}..............\left( ii \right)$

Now, let’s differentiate equation (ii) with respect to ‘x’ and using product rule of differentiation, i.e., \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], we get

$n{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right)x{{\left( 1+x \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r-1}}}$

Now multiply by ‘x’ both sides, we will get,

$nx{{\left( 1+x \right)}^{n-1}}+n\left( n-1 \right){{x}^{2}}{{\left( 1+x \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}{{x}^{r}}}.........\left( iii \right)$

Now, in the equation (iii) we will substitute ‘x’ by $\left( \dfrac{p}{q} \right)$, so we get,

$\begin{align}

& n\left( \dfrac{p}{q} \right){{\left( 1+\dfrac{p}{q} \right)}^{n-1}}+n\left( n-1 \right){{\left( \dfrac{p}{q} \right)}^{2}}{{\left( 1+\dfrac{p}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}} \\

& \Rightarrow n\left( \dfrac{p}{q} \right){{\left( \dfrac{p+q}{q} \right)}^{n-1}}+n\left( n-1 \right)\left( \dfrac{{{p}^{2}}}{{{q}^{2}}} \right){{\left( \dfrac{p+q}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}} \\

\end{align}$

In the question we were given that ‘p + q = 1’, so substituting this in above equation, we get

$n\left( \dfrac{p}{q} \right){{\left( \dfrac{1}{q} \right)}^{n-1}}+n\left( n-1 \right)\left( \dfrac{{{p}^{2}}}{{{q}^{2}}} \right){{\left( \dfrac{1}{q} \right)}^{n-2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}}$

Combining the like terms, we get

$\Rightarrow n\dfrac{p}{{{q}^{n}}}+\dfrac{n\left( n-1 \right){{p}^{2}}}{{{q}^{n}}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{\left( \dfrac{p}{q} \right)}^{r}}$

Now, multiplying ${{q}^{n}}$on both side of the above equation we get,

$np+n\left( n-1 \right){{p}^{2}}=\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}$

We can also write like this

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np+n\left( n-1 \right){{p}^{2}}$

Combining the like terms, we get

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np\left( 1+np-p \right)$

We were given that ‘p + q = 1’, so, we can replace $\left( 1-p \right)$ by $q$, the above equation can be written as,

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=np\left( q+np \right)$

Opening the bracket,w e get

$\sum\limits_{r=0}^{n}{{{r}^{2}}{}^{n}{{C}_{r}}}{{p}^{r}}{{q}^{n-r}}=npq+{{n}^{2}}{{p}^{2}}$

Hence Proved

Note: In these type of questions, student generally go wrong while differentiating;

${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}$with respect to $x$.

Another approach of this problem is

\[\sum\nolimits_{r=0}^{n}{{{r}^{2}}{{c}_{r}}{{p}^{r}}.{{q}^{n-r}}}\]

And convert this to the formula, ${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{x}^{r}}}$

In this way we can prove LHS is equal to RHS.

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