Question

For $$2 \leqslant r \leqslant n,$$ \[\left( {\begin{array}{{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right) = \]

A. \[\left( {\begin{array}{*{20}{c}} {n + 1}\\ {r - 1} \end{array}} \right)\]

B. \[2\left( {\begin{array}{*{20}{c}} {n + 1}\\ {r + 1} \end{array}} \right)\]

C. \[2\left( {\begin{array}{*{20}{c}} {n + 2}\\ r \end{array}} \right)\]

D. \[\left( {\begin{array}{*{20}{c}} {n + 2}\\ r \end{array}} \right)\]

Answer : \[\left( {\begin{array}{*{20}{c}} {n + 2}\\ r \end{array}} \right)\]

Solution :
\[\begin{array}{l} \left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right)\\ = \left[ {\left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right)} \right] + \left[ {\left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right)} \right] \end{array}\]
NOTE THIS STEP :
\[\left( {\begin{array}{*{20}{c}} {n + 1}\\ r \end{array}} \right) + \left( {\begin{array}{*{20}{c}} {n + 1}\\ {r - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {n + 2}\\ r \end{array}} \right)\]
$$\left[ {\because \,{\,^n}{C_r} + {\,^n}{C_{r - 1}} = {\,^{n + 1}}{C_r}} \right]$$

For $$2 \leqslant r \leqslant n,$$ \[\left( {\begin{array}{{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right) = \]

Releted MCQ Question on
Algebra >> Binomial Theorem

Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$ is

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

Practice More Releted MCQ Question on
Binomial Theorem

Practice More MCQ Question on Maths Section

For $$2 \leqslant r \leqslant n,$$ \[\left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right) = \]

Releted MCQ Question on Algebra >> Binomial Theorem

Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$ is

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$ is a polynomial of degree

If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is

Practice More Releted MCQ Question on Binomial Theorem

Practice More MCQ Question on Maths Section

For $$2 \leqslant r \leqslant n,$$ \[\left( {\begin{array}{{20}{c}} n\\ r \end{array}} \right) + 2\left( {\begin{array}{{20}{c}} n\\ {r - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} n\\ {r - 2} \end{array}} \right) = \]

Releted MCQ Question on
Algebra >> Binomial Theorem

Practice More Releted MCQ Question on
Binomial Theorem