Check sibling questions

Example 4 - Prove that 6^n - 5n always leaves remainder 1 when divided

Example  4 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  4 - Chapter 8 Class 11 Binomial Theorem - Part 3
Example  4 - Chapter 8 Class 11 Binomial Theorem - Part 4
Example  4 - Chapter 8 Class 11 Binomial Theorem - Part 5

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Transcript

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 We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = 1 ,b = 5 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 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 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 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 We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = 1 ,b = 5 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.