Top 50 Most Repeated BINOMIAL THEOREM PYQs | JEE MAINS
A curated collection of the most important questions from BINOMIAL THEOREM, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from BINOMIAL THEOREM, fully solved with step-by-step concepts to prepare for JEE MAINS.
Use $\frac{n+1}{1+\frac{1}{r}}$ to find term index; $r=5.5 \Rightarrow T_6$ is greatest....
Read Full Step-by-Step Solution →For \((a+b)^n\) with n = 6 (even), the single middle term is the \((n/2+1)\)th term: \(T_{4}=\binom{6}{3}x^{3}(2y)^{3}=20·x^{3}·8y^{3}=160x^{3}y^{3}\)...
Read Full Step-by-Step Solution →General term: $T_{r+1} = \binom{6}{r}(2x^2)^{6-r}\left(-\frac{1}{x}\right)^r$. Set power of $x$ to 0....
Read Full Step-by-Step Solution →Using binomial theorem, coefficient of $x^3$ is $\binom{5}{3} = 10$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r}(a)^{n-r}(b)^r$ with $r=3$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ with $r=3$....
Read Full Step-by-Step Solution →Find $r$ where $\frac{T_{r+1}}{T_r} \geq 1$. For $x=0.5$, $r \leq 3.67$, so $r=3$. $T_4 = \binom{10}{3}(0.5)^3 = 252$....
Read Full Step-by-Step Solution →The greatest term occurs where the ratio of successive terms $T_{r+1}/T_r$ is just less than 1. Setting $\frac{T_{r+1}}{T_r}=\frac{(7-r)}{r+1}\cdot\fr...
Read Full Step-by-Step Solution →$(1+x)^n \approx 1 + nx$ for small $x$. Here $x=0.01$, $n=5$....
Read Full Step-by-Step Solution →For small $x$, $(1+x)^n \approx 1+nx$. Here $x=0.02$ and $n=5$, so $(1+0.02)^5 \approx 1+5\times0.02 = 1.10$....
Read Full Step-by-Step Solution →Set power of $x$ to zero:
Sum of binomial coefficients in row $n$ is
This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Binomial Theorem....
Read Full Step-by-Step Solution →Use $\frac{n+1}{1+\frac{1}{r}}$ to find term index; $r=5.5 \Rightarrow T_6$ is greatest....
Read Full Step-by-Step Solution →For \((a+b)^n\) with n = 6 (even), the single middle term is the \((n/2+1)\)th term: \(T_{4}=\binom{6}{3}x^{3}(2y)^{3}=20·x^{3}·8y^{3}=160x^{3}y^{3}\)...
Read Full Step-by-Step Solution →General term: $T_{r+1} = \binom{6}{r}(2x^2)^{6-r}\left(-\frac{1}{x}\right)^r$. Set power of $x$ to 0....
Read Full Step-by-Step Solution →Using binomial theorem, coefficient of $x^3$ is $\binom{5}{3} = 10$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r}(a)^{n-r}(b)^r$ with $r=3$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ with $r=3$....
Read Full Step-by-Step Solution →Find $r$ where $\frac{T_{r+1}}{T_r} \geq 1$. For $x=0.5$, $r \leq 3.67$, so $r=3$. $T_4 = \binom{10}{3}(0.5)^3 = 252$....
Read Full Step-by-Step Solution →The greatest term occurs where the ratio of successive terms $T_{r+1}/T_r$ is just less than 1. Setting $\frac{T_{r+1}}{T_r}=\frac{(7-r)}{r+1}\cdot\fr...
Read Full Step-by-Step Solution →$(1+x)^n \approx 1 + nx$ for small $x$. Here $x=0.01$, $n=5$....
Read Full Step-by-Step Solution →For small $x$, $(1+x)^n \approx 1+nx$. Here $x=0.02$ and $n=5$, so $(1+0.02)^5 \approx 1+5\times0.02 = 1.10$....
Read Full Step-by-Step Solution →Set power of $x$ to zero:
Sum of binomial coefficients in row $n$ is
This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Binomial Theorem....
Read Full Step-by-Step Solution →Use $\frac{n+1}{1+\frac{1}{r}}$ to find term index; $r=5.5 \Rightarrow T_6$ is greatest....
Read Full Step-by-Step Solution →For \((a+b)^n\) with n = 6 (even), the single middle term is the \((n/2+1)\)th term: \(T_{4}=\binom{6}{3}x^{3}(2y)^{3}=20·x^{3}·8y^{3}=160x^{3}y^{3}\)...
Read Full Step-by-Step Solution →General term: $T_{r+1} = \binom{6}{r}(2x^2)^{6-r}\left(-\frac{1}{x}\right)^r$. Set power of $x$ to 0....
Read Full Step-by-Step Solution →Using binomial theorem, coefficient of $x^3$ is $\binom{5}{3} = 10$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r}(a)^{n-r}(b)^r$ with $r=3$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ with $r=3$....
Read Full Step-by-Step Solution →Find $r$ where $\frac{T_{r+1}}{T_r} \geq 1$. For $x=0.5$, $r \leq 3.67$, so $r=3$. $T_4 = \binom{10}{3}(0.5)^3 = 252$....
Read Full Step-by-Step Solution →The greatest term occurs where the ratio of successive terms $T_{r+1}/T_r$ is just less than 1. Setting $\frac{T_{r+1}}{T_r}=\frac{(7-r)}{r+1}\cdot\fr...
Read Full Step-by-Step Solution →$(1+x)^n \approx 1 + nx$ for small $x$. Here $x=0.01$, $n=5$....
Read Full Step-by-Step Solution →For small $x$, $(1+x)^n \approx 1+nx$. Here $x=0.02$ and $n=5$, so $(1+0.02)^5 \approx 1+5\times0.02 = 1.10$....
Read Full Step-by-Step Solution →Set power of $x$ to zero:
Sum of binomial coefficients in row $n$ is
This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Binomial Theorem....
Read Full Step-by-Step Solution →Use $\frac{n+1}{1+\frac{1}{r}}$ to find term index; $r=5.5 \Rightarrow T_6$ is greatest....
Read Full Step-by-Step Solution →For \((a+b)^n\) with n = 6 (even), the single middle term is the \((n/2+1)\)th term: \(T_{4}=\binom{6}{3}x^{3}(2y)^{3}=20·x^{3}·8y^{3}=160x^{3}y^{3}\)...
Read Full Step-by-Step Solution →General term: $T_{r+1} = \binom{6}{r}(2x^2)^{6-r}\left(-\frac{1}{x}\right)^r$. Set power of $x$ to 0....
Read Full Step-by-Step Solution →Using binomial theorem, coefficient of $x^3$ is $\binom{5}{3} = 10$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r}(a)^{n-r}(b)^r$ with $r=3$....
Read Full Step-by-Step Solution →Use general term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ with $r=3$....
Read Full Step-by-Step Solution →Find $r$ where $\frac{T_{r+1}}{T_r} \geq 1$. For $x=0.5$, $r \leq 3.67$, so $r=3$. $T_4 = \binom{10}{3}(0.5)^3 = 252$....
Read Full Step-by-Step Solution →The greatest term occurs where the ratio of successive terms $T_{r+1}/T_r$ is just less than 1. Setting $\frac{T_{r+1}}{T_r}=\frac{(7-r)}{r+1}\cdot\fr...
Read Full Step-by-Step Solution →