Example 4 - Chapter 8 Class 11 Binomial Theorem

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Example 4 (Introduction) Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25. Taking examples ﷐26﷮25﷯ = 1﷐1﷮25﷯ we can write 26 as 25 × 1 + 1 ﷐51﷮25﷯ = 2﷐1﷮25﷯ we can write 51 as 25 × 2 + 1 ﷐251﷮25﷯ = 10﷐1﷮25﷯ 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