Wanna practice Induction ?

The principle of mathematical induction is a very very useful tool in many proofs, and also in proving some useful formulas. The principle states that any given set of positive integers\color{#D61F06}{\text{The principle states that any given set of positive integers}} has ALL natural numbers if it follows the following conditions -

(i)(i) The first natural number, i.e. 11 is in the set.

(ii)(ii) Whenever the integer kk is in the set, then k+1k+1 is also in the set.

This looks so obvious, see that if you have been given the two statements simultaneously, then by using (ii)(ii) on (i)(i), you can say that 22 is in the set. Then again applying the statement (ii)(ii) for 22, we say that (3)(3) will be in the set, so on.

Thus if you prove a certain result for 1\color{#3D99F6}{\textbf{Thus if you prove a certain result for 1}} , and then assume that the result is true\color{#3D99F6}{\textbf{and then assume that the result is true}} for kk, just prove that it is true for k+1k+1 and you're done !

Same thing you can do for 0, then prove for whole numbers !!!


Problems for practice:-

Prove that the following results hold nN\forall n \in \mathbb{N}

(a)k=0n2k=2n+11(a) \displaystyle \sum_{k=0} ^n 2^k = 2^{n+1}-1

(b)k=1nk=n(n+1)2(b) \displaystyle \sum_{k=1} ^n k = \dfrac{n(n+1)}{2}

(c)k=1nk2=n(n+1)(2n+1)6(c) \displaystyle \sum_{k=1} ^n k^2 = \dfrac{n(n+1)(2n+1)}{6}

(d)k=1nk3=(k=0nk)2(d) \displaystyle \sum_{k=1}^n k^3 = \biggl( \sum_{k=0} ^n k \biggr) ^2

(e)2427n+35n5(e) \displaystyle 24 \mid 2\cdot 7^n + 3\cdot 5^n -5

(f)12+23+34+....+n(n+1)=n(n+1)(n+2)3(f) \displaystyle 1\cdot 2+2\cdot 3+3\cdot 4+ .... + n(n+1) = \dfrac{n(n+1)(n+2)}{3}

I felt like sharing because though it's very well known to all, I am willing to find some good level problems for practicing this foundation builder concept once more... Isn’t this a good revision ?\color{#20A900}{\text{Isn't this a good revision ?}}

Note by Aditya Raut
5 years, 1 month ago

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@Aditya Raut this is really a nice note for practice...... could you please add more divisibility kinda problems to give a more diverse experience to readers? ........ thanks :)

Abhinav Raichur - 5 years, 1 month ago

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Here is some additional problems I found on the internet: http://staffhome.ecm.uwa.edu.au/~00021149/Academy/1995/inductionprobs.pdf

Daniel Liu - 5 years, 1 month ago

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Thanks - they look like a good selection of induction problems

Curtis Clement - 4 years, 8 months ago

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I actually haven't tried (d)(d) and (f)(f) before since doing math XD

Here's the solution for (e)(e) if someone got stuck.

2×7m+1+3×5m+152\times 7^{m+1} + 3 \times 5^{m+1} - 5

=2×7m×7+3×5m×55= 2\times 7^{m}\times 7 + 3\times 5^{m}\times 5 - 5

=14×7m+15×5m5= 14\times 7^{m} + 15 \times 5^{m} - 5

=12×7m+12×5m+2×7m+3×5m5= 12\times 7^{m} + 12\times 5^{m} + 2\times 7^{m} + 3\times 5^{m} - 5

=12(7m+5m)+24k = 12(7^{m} + 5^{m}) + 24k for some kk

=24n+24k = 24n + 24k for some nn

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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(a)(a) Base Case n=1n = 1

k=012k=3\displaystyle \sum_{k = 0}^{1}2^{k} = 3

2n+11=32^{n+1} - 1 = 3

True.

Propose it is true nN\forall n \in N

Then it must be true for n+1n + 1

k=0n+12k=k=0n2k+2n+1\displaystyle\sum_{k = 0}^{n+1}2^{k} = \sum_{k = 0}^{n}2^{k} + 2^{n+1}

Substituting the value from our proposal,

k=0n2k+2n+1=2n+11+2n+1=2n+21\sum_{k = 0}^{n}2^{k} + 2^{n+1} = 2^{n+1} - 1 + 2^{n+1}= 2^{n+2} - 1

Thus, LHS=RHSLHS = RHS

Thus, LHS=RHSnNLHS = RHS \forall n \in N

Sorry for ugly arrangement.

Sanchayapol Lewgasamsarn - 5 years, 1 month ago

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@Aditya Raut Can you add these to the Induction Wiki page? Thanks!

Calvin Lin Staff - 4 years, 10 months ago

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Nice! I'm too into creating notes and problems set for inequality :) Why not take a look at it @Aditya Raut ? :D

Priyansh Sangule - 5 years, 1 month ago

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What is e) ?... in d) second summation is it k=0 or k=1 ? Nice collection.

Niranjan Khanderia - 4 years, 10 months ago

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Thanks, but Just think ! Will adding 00 make any change? In that summation, starting from k=0k=0 will yield same thing as starting with k=1k=1, won't it ? @Niranjan Khanderia

Aditya Raut - 4 years, 10 months ago

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Thanks.

Niranjan Khanderia - 4 years, 10 months ago

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And e) means prove that for all positive integers nn, (27n+35n5)(2\cdot 7^n + 3\cdot 5^n -5) is divisible by 2424.

aba\mid b is the symbol of "a divides b" or "b is divisible by a".

Latex code for the symbol \mid is "\mid"

Aditya Raut - 4 years, 10 months ago

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Thanks. I did not get it since I did not assume the parenthesis.

Niranjan Khanderia - 4 years, 10 months ago

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/(good/)

samuel ayinde - 4 years, 7 months ago

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A good level problem is here: Fro every positive integer nn, prove that (4n+1)<n+(n+1)<(4n+2)\sqrt{(4n+1)}<\sqrt n +\sqrt{(n+1)}<\sqrt{(4n+2)}. Hence or otherwise, prove that [n+n+1\sqrt{n}+\sqrt{n+1}]=[4n+1\sqrt{4n+1}] where [.] denotes floor function. It is an IITJEE problem.

Gautam Sharma - 4 years, 4 months ago

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