Mindblank on this induction q (1 Viewer)

medaspirant · Dec 29, 2021

part a idk i just cant seem to figure it out any guidance is appreciated

Trebla · Dec 29, 2021

From the inductive step
$1^3+2^3+3^3+...+k^3+(k+1)^3\\\\=\dfrac{k^2(k+1)^2}{4}+(k+1)^3\\\\=\dfrac{(k+1)^2(k^2+4(k+1))}{4}\\\\=\dfrac{(k+1)^2(k+2)^2}{4}$

yanujw · Dec 29, 2021

To find the limit in (a)
$\frac{}{}1^3 + 2^3 + 3^3 ... + n^3 = \frac{n^2(n+1)^2}{4}\\ \frac{1^3 + 2^3 + 3^3 ... + n^3}{n^4} = \frac{n^2(n+1)^2}{4n^4}$
$=\displaystyle \lim_{n \to infinity}\frac{n^2+2n+1}{4n^2}$
$=\displaystyle \lim_{n \to infinity}\frac{1}{4}+\frac{1}{2n}+\frac{1}{4n^2}$
The two last fractions are hyperbolas, they approach 0 as x approaches infinity.
Therefore, the limit is $\frac{1}{4}$

5uckerberg · Dec 30, 2021

medaspirant said:
View attachment 34524
part a idk i just cant seem to figure it out any guidance is appreciated

For part b you are going to have to use the fact that $1+2+3+4+...+n=\frac{n(n+1)}{2}$ along your line of working to complete the question.

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