Question

Using binomial theorem prove that 2^3n-7n-1 is divisible by 49

Answer 1

Answer:

The proof is explained below :

Step-by-step explanation:

$=2^{3\cdot n}-7\cdot n-1\\=8\cdot n-7\cdot n-1\\=(7+1)^n-7\cdot n-1\\\\=\thinspace _{0}^{n}\textrm{C}\cdot 7^n\cdot 1^0 +\thinspace _{1}^{n}\textrm{C}\cdot 7^{n-1}\cdot 1^1 +\thinspace _{2}^{n}\textrm{C}\cdot 7^{n-2}\cdot 1^2 +............+\thinspace _{n-2}^{n}\textrm{C}\cdot 7^2\cdot 1^{n-2}+\\\\\thinspace _{n-1}^{n}\textrm{C}\cdot 7^1\cdot 1^{n-1} +\thinspace _{n}^{n}\textrm{C}\cdot 7^0\cdot 1^n -7\cdot n -1\\\\\text{Now upto }7^{2}\text{ , 49 can be taken out common from each term and the remaining terms}\\\\\text{ can be considered as a constant m}\\\\=49\cdot m+\frac{n!\cdot 7}{(n-1)!}+1-7\cdot n-1\\\\=49\cdot m +n\cdot 7-7\cdot n\\\\=49\cdot m$

which clearly shows the given expression is divisible by 49.

Hence Proved.

Answer 2

Hey mate! Your answer is in the attachment