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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
Show that $n^3-n$ is divisible by $6$ using induction This answer is with basic induction method when n=1, $\ 1^3-1 = 0 = 6 0$ is divided by 6 so when n=1,the answer is correct we assume that when n=p , the answer is correct so we take, $\ p^3-p $ is divided by 6 then, when n= (p+1), $$\ (p+1)^3- (p+1) = (P^3+3p^2+3p+1)- (p+1)$$ $$\ =p^3-p+3p^2+3p+1-1 $$ $$\ = (p^3-p)+3p^2+3p $$ $$\ = (p^3-p)+3p (p+1) $$ as we assumed $\ (p^3-p) $ is
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
Prove that $2^n3^ {2n}-1$ is always divisible by 17 7 Prove that $2^n3^ {2n} -1$ is always divisible by $17$ 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