Question

Using binomial theorem, prove that $6^n-5n$ always leaves remains $1$ when divided by $25$.

Answer 1

$(a+b{)}^n{=}^n{c}_0{a}^n{b}^0{+}^n{c}_1{a}^{n+1}{b}^1{+}^n{c}_2{a}^{n+2}{b}^2+....{+}^n{c}_n{a}^0{b}^n$

$6^n=(1+5{)}^n{=}^n{c}_0{.1}^n{.5}^0{+}^n{c}_1{.1}^{n-1}{.5}^1{+}^n{c}_2{.1}^{n-2}{.5}^2+......{+}^n{c}_n{.1}^0{.5}^n$

$=1+n.5+\frac{n(n-1)}{2}{.5}^2+....+{1.5}^n$

$6^n=1+5n+\frac{n(n-1)}{2}{.5}^2+......+5^n$

$6^n-5n=1+\frac{n(n-1)}{2}{.5}^2+......+5^n$

$6-5n^n=1+5^2[\frac{n(n-1)}{2}.+......+5^{n-2}]$

$6-5n^n=1+25[\frac{n(n-1)}{2}.+......+5^{n-2}]$

Let $\frac{n(n-1)}{2}+....+5^{n-2}=K$ (some positive integer)

$6^n-5n=1+25K$

i.e., Remainder is always $1$ here $|$