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- Prove that $1^3 + 2^3 + . . . + n^3 = (1+ 2 + . . . + n)^2$
Thus by the principle of mathematical induction 13 +23 +33 + ⋯ +n3 = (1 + 2 + 3 + ⋯ + n)2 1 3 + 2 3 + 3 3 + + n 3 = (1 + 2 + 3 + + n) 2 for each n ∈N n ∈ N
- recursive algorithms - Recursion tree T (n) = T (n 3) + T (2n 3) + cn . . .
Recursion tree for T(n) = T(n3) + T(2n3) + cn T (n) = T (n 3) + T (2 n 3) + c n Shortest path will be most left one, because it operates on lowest value, and the most right one will be the longest one, that means tree is not balanced
- Proving by induction: $2^n gt; n^3 - Mathematics Stack Exchange
I need to prove that 2n>n3 ∀n ∈N, n> 9 2 n> n 3 ∀ n ∈ N, n> 9 Now that is actually very easy if we prove it for real numbers using calculus But I need a proof that uses mathematical induction I tried the problem for a long time, but got stuck at one step - I have to prove that: k3> 3k2 + 3k + 1 k 3> 3 k 2 + 3 k + 1 Hints???
- divisibility - Proving $n^3 + 3n^2 +2n$ is divisible by $6 . . .
The full question is: Factorise n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n Hence prove that when n n is a positive integer, n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n is always divisible by 6 6 So i factorised and got n(n + 1)(n + 2) n (n + 1) (n + 2) which i think is right? I'm not sure how to actually prove this is divisible by 6 6 though Thanks for help and i apologise if someone has already asked this, i
- Big-O Notation - Prove that $n^2 - Mathematics Stack Exchange
Basically: they did it because it was easy! The real idea of Big-O notation is to find whatever term gives you the major contribution -- in this case, we know that x2 x 2 is much larger than x x when x x is large -- and bound by it They could just as easily have said that when x ≥ 2 x ≥ 2, we have 2x ≤x2 2 x ≤ x 2 and 1 ≤x2 1 ≤ x 2, and made the constant 3 The specifics can vary
- Prove that $2^n3^{2n}-1$ is always divisible by 17
Prove that 2n32n − 1 2 n 3 2 n 1 is always divisible by 17 17 I am very new to proofs and i was considering using proof by induction but I am not sure how to I know you have to start by verifying the statement is true for the integer 1 but I dont know where to go from there
- Show that $f(n)=n^3+20n+1=O(n^3)$ - Mathematics Stack Exchange
If it were O(n2) O (n 2) each of n3 n 3, 20n 20 n and 1 1 would have to be less than Cn2 C n 2 for a fixed C ∈R+ C ∈ R + for all n n to infinity Clearly this is impossible for n3 n 3
- sequences and series - Does $\sum_ {n=1}^ {\infty} (n^3 +1 )^ {1 3}-n . . .
Suppose I am given a infinite series as $$\sum_ {n=1}^ {\infty} (n^3 +1 )^ {1 3}-n$$ how can I tell that if it converges or diverges (by which test) , I applied D'alembert ratio test as $$\lim_ {n \to
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