Prove that $\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$.
(N/A) By the Binomial Theorem,we have the expansion:
$\sum\limits_{r = 0}^n {^nC_r a^{n - r} b^r} = (a + b)^n$
By substituting $a = 1$ and $b = 3$ in the above equation,we obtain:
$\sum\limits_{r = 0}^n {^nC_r (1)^{n - r} (3)^r} = (1 + 3)^n$
Since $(1)^{n - r} = 1$,the expression simplifies to:
$\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$
Hence,the identity is proved.
$\sum\limits_{r = 0}^n {^nC_r a^{n - r} b^r} = (a + b)^n$
By substituting $a = 1$ and $b = 3$ in the above equation,we obtain:
$\sum\limits_{r = 0}^n {^nC_r (1)^{n - r} (3)^r} = (1 + 3)^n$
Since $(1)^{n - r} = 1$,the expression simplifies to:
$\sum\limits_{r = 0}^n {3^r \,^nC_r} = 4^n$
Hence,the identity is proved.
Explore More
Similar Questions
If $(3+\sqrt{2})^6-(3-\sqrt{2})^6=a+b \sqrt{2}$,then $a+b=$
The sum of coefficients in $(1 + x - 3x^2)^{2134}$ is
Easy
View SolutionLet $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $(x+\sqrt{x^3-1})^5+(x-\sqrt{x^3-1})^5, x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$,then $u+v$ equals:
DifficultJEE MAIN 2025
View SolutionThe coefficient of $x^6$ in the power series expansion of $\frac{x^4-12x^2+7}{(x^2+1)^3}$ is
If the sum of the coefficients in the expansion of $({\alpha ^2}{x^2} - 2\alpha x + 1)^{51}$ vanishes,then the value of $\alpha$ is
MediumIIT 1991
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo