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
•
Total views: 305.7k
•
Views today: 2.85k
Answer
305.7k+ views
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
