Question

test

Answer 1

ouf

Answer 2

Summation notation, looks like test passed 👌🏻

Answer 3

you got lucky pal: 5/10 is an act of extreme charity from your instructor for "that" effort :) I mean, Yuk.

... though your instructor deserves something like 1/10 for poor instruction

... as you clearly haven't a clue what you are doing, no offense intended

... and she/he is positively enforcing that approach :( by passing you when....

Perhaps you don't care, but I will post a road map. this is a **fun** problem if you are interested in using your brain as it was **never** intended to be used.

Answer 4

\(\sum\limits_{i= 1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\)

Assume true for \(n = k\):

\(P(k): \qquad \sum\limits_{i= 1}^k i^2 = \dfrac{k(k+1)(2k+1)}{6}\)

\(\therefore \sum\limits_{i= 1}^{\color{red}{k+1}} i^2 = \dfrac{k(k+1)(2k+1)}{6} + \color{red}{(k+1)^2}\)

factor out k+1


\( \qquad = (k+1) \cdot \dfrac{k (2k+1) + 6 k + 6}{6} \)

\( \qquad = \dfrac{(k+1) (2k^2+ 7 k + 6)}{6} \)

\( \qquad = \dfrac{(k+1) (k + 2) (2 k + 3) }{6} \)

\( \qquad = \dfrac{(k+1) ((k+1) + 1) (2 (k+1) + 1) }{6} \)

To make it totally clear what has happened, let \(\Omega = k + 1\)

\( \qquad = \dfrac{\Omega (\Omega + 1) (2 \Omega + 1) }{6} \)

\(\qquad = P (\Omega ) \qquad \equiv P(k+1)\) !!


You did the base case already, for whiich you scored 5 out of smudge




So just finish it :)

Answer 5

In your screen grab. More Red than Black. Mental Teacher.