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logic - 3 Knights and Knaves - Mathematics Stack Exchange So N3 can,t be a knight so he is also a knave and,we have 2 knaves,but we don,t know if the N2 is a knave or not but if N2 was a knave,N1 would be a knight which leads to a contradiction, so N2 is a knight and this completely make sense
Proving $1^3+ 2^3 + \cdots + n^3 = \left (\frac {n (n+1)} {2}\right)^2 . . . Hint $ $ First trivially inductively prove the Fundamental Theorem of Difference Calculus $$\rm\ F (n) = \sum_ {k\, =\, 1}^n f (k)\, \iff\, F (n) - F (n\!-\!1)\, =\, f (n),\ \ \, F (0) = 0\qquad$$ The result now follows immediately by $\rm\ F (n) = (n\: (n\!+\!1) 2)^2\ \Rightarrow\ \color {#c00} {F (n)-F (n\!-\!1) = n^3}$ The theorem reduces the proof to a trivial mechanical verification of a
summation - Prove that $1^3 + 2^3 + . . . + n^3 = (1+ 2 + . . . + n)^2 . . . HINT: You want that last expression to turn out to be $\big (1+2+\ldots+k+ (k+1)\big)^2$, so you want $ (k+1)^3$ to be equal to the difference $$\big (1+2+\ldots+k+ (k+1)\big)^2- (1+2+\ldots+k)^2\; $$ That’s a difference of two squares, so you can factor it as $$ (k+1)\Big (2 (1+2+\ldots+k)+ (k+1)\Big)\; \tag {1}$$ To show that $ (1)$ is just a fancy way of writing $ (k+1)^3$, you need to
For $f(n)$ find a simple $g(n)$ such that $f(n)=\\Theta(g(n))$ $$\lim_ {n\to \infty}\sum_ {i=1}^n3 (4^i)+3 (3^i)-i^ {19}+20 \to \int ?$$ I know some basics of Riemann's transition from sum to integral with dx, but don't know what to do in this case