Question bank/JEE Main/Mathematics
JEE MainMathematicsToughBinomial Theorem
Suppose [IMAGE] holds for some positive integer n. Then [IMAGE] equals.
Show answer and solution
$\left| \begin{matrix}
\frac{n(n + 1)}{2} & n2^{n - 1} + n(n - 1)2^{n - 2} \\
n2^{n - 1} & 4^{n}
\end{matrix} \right| = 0$
n = 0 or 4(n + 1) - 2n - n(n - 1) = 0
4n + 4 - 2n - n^2 + n = 0
3n - n^2 + 4 = 0 n^2 - 3n - 4 = 0
(n - 4)(n + 1) = 0
n = 4
\frac{n(n + 1)}{2} & n2^{n - 1} + n(n - 1)2^{n - 2} \\
n2^{n - 1} & 4^{n}
\end{matrix} \right| = 0$
n = 0 or 4(n + 1) - 2n - n(n - 1) = 0
4n + 4 - 2n - n^2 + n = 0
3n - n^2 + 4 = 0 n^2 - 3n - 4 = 0
(n - 4)(n + 1) = 0
n = 4
Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx
Need a video for this idea?
Scan any doubt from your notes and get a whiteboard explanation.