Question

Using the binomial theorem ,prove that ( 6ⁿ-5n) always leaves the remainder 1 when divided by 25

Answer 1

we can write (6ⁿ ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! +.............∞ use this here,

(6)ⁿ = (1 + 5)ⁿ = 1 + 5n + n(n-1)5²/2! + n(n-1)(n-2)5³/3! +...........∞

= 1 + 5n + 5²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}

Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞

6ⁿ = 1 + 5n + 25P
6ⁿ - 5n = 1 + 25P -------(1)

but we know, according to Euclid algorithm ,
dividend = divisor × quotient + remainder ---(2)
compare eqn (1) to (2)

we observed that 6ⁿ -5 n always leaves the remainder 1 when divided by 25

Answer 2

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THANK YOU.