# Example 4 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 4 (Introduction) Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25. Taking examples 2625 = 1125 we can write 26 as 25 × 1 + 1 5125 = 2125 we can write 51 as 25 × 2 + 1 25125 = 10125 we can write 251 as 25 × 10 + 1 Any number which leaves remainder 1 when divided by 25 = 25 × Natural number + 1 Example 4 Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25. Writing 6n = (1 + 5)n (6)n = nC0 1n 50 + nC1 1n – 1 51 + nC2 1n – 2 52…... + nCn 1n – n 5n = nC0 50 + nC1 51 + nC2 52 ………..…... + nCn 5n = 1 × 1 + 𝑛!1! 𝑛 −1! 51 + 𝑛!2! 𝑛 −2! 52………. + 1 × 5n = 1 + 𝑛(𝑛−1)!1 𝑛−1! 5 + 𝑛(𝑛 − 1)(𝑛 − 2)!2 𝑛 −2! 52………. + 5n = 1 + n(5) + 𝑛(𝑛 − 1) 2 52………. + 5n Thus, (6)n = 1 + 5n + 𝑛(𝑛 − 1) 2 52………. + 5n 6n – 5n = 1 + 𝑛(𝑛 − 1) 2 52………. + 5n 6n – 5n = 1 + 52 𝑛(𝑛 − 1) 2 ………. + 5n – 2 6n – 5n = 1 + 25 𝑛(𝑛 − 1) 2 ………. + 5n – 2 6n – 5n = 1 + 25k where k = ( 𝑛(𝑛 − 1) 2 ………. + 5n – 2 ) The above equation is of the form Dividend = Divisor × Quotient + Remainder Putting Dividend = 6n – 5n , Divisor = 25 , Remainder = 1 6n – 5n = 25k + 1 Hence 6n – 5n always leave Remainder 1 when dividing by 25 Hence Proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.