Prove that 1+3+5+...+(2n-1)=n2 for every positive integer n by adding 1+3+5+...+(2n-1) and (2n-1)+(2n-3)+...+1?
1+3+5+...+(2n-1) = S (2n-1)+(2n-3)+...+1=S Those two are the same sum, the second in reverse order. Add vertically and get: 2n+2n+2n+...+2n There are n terms, so you get 2S= n(2n) = 2n^2. Divide by 2: S=n^2.
Use induction to prove that 1 + 2 + 5 +lll+ (2n-1) = N^2 for all positive integers n?
A small point - the second term of your series should be 3, not ; also the parameter n or N should be consistent on both sides ie use one not both In the induction method, we assume that the expression 1 + 3 + ... (2.N - 1) = N^2 is correct for a given value of N, and demonstrate that the next value is also correct 1 + 3 + ... (2.N - 1) + [2(N + 1) - 1] = N^2 + [2(N + 1) - 1] = N^2 + 2.N + 1 = (N + 1)^2 To complete the proof, we show that the expression is correct for one or more particular values of N eg for N = 2 and 3 1 + 3 = 4 = 2^2 1 + 3 + 5 = 9 = 3^2 QED
Prove that if n is a positive integer then 13 | 4^(2n−1) + 3^(n+1)?
Multiply by the right side by 3^(2n-1) to get 12^(2n-1)+3^(3n) 12= - 1 mod13 and 3^3= +1 put these in: (-1)^(2n-1) +1^(3n) = 0 mod 13. Therefore, 13 | 4^(2n-1)+3^(n+1).
How can you prove by induction that 7^n+ 4^n + 1 is divisible by 6, for n = 1, 2, 3, …?
The question can be presented as:Prove that 6|7^n+4^n+1 when n is positive intergerLet P(n) be the proposition that asserts 7^n+4^n+1 is divisible by 6.Basic step:P(1) is true, since 7+4+1 = 12, 6|12Inductive step:Assume that P(k) is true, with k>=1.That is:6|7^k+4^k+1 We need to prove P(k+1) is true:That is 7^(k+1)+4^(k+1)+1 = 7*7^(k) + 4*4^(k)+1=(6+1)*7^(k) + (3+1)*4^(k)+1=(7^k+4^k+1 ) + (6*7^k + 3*4^k)two terms are divisible by 6, so P(k+1) is true.This completes the proof
How can you prove that 4^n - 1 is divisible by 3 using induction?
Trivial case: [math]n=0[/math]. Observe that [math]4^0–1[/math] is divisible by [math]3[/math]. Done.Inductive case: Assume [math]4^n-1[/math] is divisible by [math]3[/math]. Show that this implies [math]4^{n+1}-1[/math] is divisible by [math]3[/math].[math]4^{n+1}-1 = 4\cdot4^n -1[/math][math]= (3+1)4^n -1[/math][math]= 3\cdot 4^{n} + 4^n - 1[/math][math]= (3\cdot 4^{n}) + (4^n - 1)[/math]The first term is divisible by [math]3[/math], since it has three as a factor.The second term is divisible by [math]3[/math] by the inductive hypothesis.So the whole thing is divisible by [math]3[/math].By induction,[math]4^n-1[/math] is divisible by [math]3[/math] for any nonnegative integer [math]n[/math].
Prove (2n+1) + (2n+3) + (2n+5) + ... + (4n-1) = 3(n^2) for all positive integers n?
You can use mathematical induction. At n = 1, LHS = 3 RHS = 3 So, it works. Assume at n = k, (2k+1) + (2k+3) + (2k+5) + ... + (4k-1) = 3(k^2) At n = k+1 (2k+3) + (2k+7) + (2k+9) + ... + (4k+3) = 3(k^2) - (2k+3) + (4k+1) + (4k+3) = 3k^2 + 6k +1 = 3(k^2+2k+1) = 3(k+1)^2 QED
Prove by Induction: 1 +5+9...+(4n -3) = n(2n -1)?
1 + 5 + 9 + ... + 4(n-3) = n(2n-1) Is it true for n = 1? 1 =?= 1(2(1) - 1) = 1. True for n = 1 Is it true for n, if it is true for n-1? 1 + 5 + 9 + ... + 4((n-1) - 3) = (n-1)(2(n-1) - 1). By assumption - because we assume it is true for the case n-1. Expand right-hand-side: (n-1) (2 (n-1) - 1) = n (2(n-1) - 1) - (2(n-1) - 1) = n (2n - 3) - (2n - 2) Add 4n - 3, rearrange terms, and you will get: = n (2n - 1)
How can using mathematical induction prove that 21 divides [math]4^{n+1}+ 5^{2n-1}?[/math]
Let [math]S(n)[/math] be the statement: [math]4^{n+1}+5^{2n-1}[/math] is divisible by [math]21[/math]Basis step: [math]S(1)[/math]: [math]4^{(1)+1}+5^{2(1)-1}[/math][math]\hspace{26 mm}=4^{2}+5^{1}[/math][math]\hspace{26 mm}=16+5[/math][math]\hspace{26 mm}=21[/math], which is divisible by [math]21[/math]Inductive step:Assume [math]S(k)[/math] is true, i.e. assume that [math]4^{k+1}+5^{2k-1}[/math] is divisible by [math]21[/math][math]\hspace{60 mm}\Rightarrow 4^{k+1}+5^{2k-1}=21A[/math]; [math]A\in\mathbb{N}[/math][math]S(k+1)[/math]: [math]4^{(k+1)+1}+5^{2(k+1)-1}[/math][math]\hspace{14 mm}=4^{k+2}+5^{2k+2-1}[/math][math]\hspace{14 mm}=4^{k+2}+5^{2k+1}[/math][math]\hspace{14 mm}=4\bullet{4^{k+1}}+25\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=4\bullet{4^{k+1}}+(21+4)\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=4\bullet{4^{k+1}}+21\bullet{5^{2k-1}}+4\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=4\hspace{1 mm}(4^{k+1}+5^{2k-1})+21\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=4\bullet{21A}+21\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=21\bullet{4A}+21\bullet{5^{2k-1}}[/math][math]\hspace{14 mm}=21\hspace{1 mm}(4A+5^{2k-1})[/math], which is divisible by [math]21[/math]So, [math]S(k+1)[/math] is true whenever [math]S(k)[/math] is true.Therefore, [math]4^{n+1}+5^{2n-1}[/math] is divisible by [math]21[/math].
How do I prove that (n^3) +2n is divisible by 3 for some integer n?
Let’s prove it using principle of mathematical induction(PMI).P(n)=n^3+2nNow for n=1,P(1)=1+2=3 which is divisible by 3.Now for n=k,P(k)=k^3+2kAssume P(k) be divisible by 3.Now for n=k+1,P(k+1)=(k+1)^3+2(k+1)P(k+1)=k^3+2k+3k^2+3k+3=P(k)+3(k^2+k+1)Since we assumed P(k) to be divisible by 3 therefore P(k+1) is also divisible by 3.Hence by PMI n^3+2n is divisible by 3 for some integer n.