Question

Stuck on series/sequences question. *see below*

Answer 1

So I've already done part (i) and ended up with 3, 4, 6, 9 and I'm on to part (ii) with PMI and I'm having some difficulty with proving that the statement is true for n=k+1. I have
\[u _{k+1} = \frac{ k ^{2}-k+6 }{ 2 } +k\]

Answer 2

I'm wondering if this is correct?

Answer 3

yes, add the fractions and complete the square for numerator

Answer 4

Thank you :) I've just realized that ths will give me the same result as \[u _{k+1}=\left( (k+1)^{2}-(k+1)+6 \right)\]

Answer 5

all divided by 2

Answer 6

\[\begin{align}u _{k+1} &= \frac{ k ^{2}-k+6 }{ 2 } +k\\~\\
&= \frac{ k ^{2}-k+6 }{ 2 } +\dfrac{2k}{2}\\~\\
&= \frac{ k ^{2}+2k+1-k+5 }{ 2 } \\~\\
&= \frac{ (k+1)^2-(k+1)+6 }{ 2 } \\~\\

\end{align}\]

Answer 7

yup :)

Answer 8

Wooow! :D

Answer 9

Thank you :)

Answer 10

yw:)