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a simple bit of simplifying

I'm doing a proof by induction question for FP2 and I can't simplify the bit at the end. It's just indices. I'm kinda embarrassed.:colondollar:

how do I simplify:

[(2^k)k] + [(k+2)^(2k+1)]
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by LauraDG
I'm doing a proof by induction question for FP2 and I can't simplify the bit at the end. It's just indices. I'm kinda embarrassed.:colondollar:

how do I simplify:

[(2^k)k] + [(k+2)^(2k+1)]


Is that 2kk+(k+2)2k+12^kk + (k+2)^{2k+1}
Reply 2
Avatar for LauraDG
LauraDG
OP
Original post by TenOfThem
Is that 2kk+(k+2)2k+12^kk + (k+2)^{2k+1}


Yes it is :smile:
Reply 3
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TenOfThem
18
Original post by LauraDG
Yes it is :smile:


Well you can split the (k+2)2k+1(k+2)^{2k+1} up a bit

(k+2)(k+2)2k(k+2)(k+2)^{2k}

(k+2)(k2+4k+4)k(k+2)(k^2+4k+4)^k

Which can then split into a selection of terms

I do not really like this TBH

Not knowing what you are aiming for does not help :s-smilie:
Reply 4
Avatar for LauraDG
LauraDG
OP
Original post by TenOfThem
Well you can split the (k+2)2k+1(k+2)^{2k+1} up a bit

(k+2)(k+2)2k(k+2)(k+2)^{2k}

(k+2)(k2+4k+4)k(k+2)(k^2+4k+4)^k

Which can then split into a selection of terms

Not knowing what you are aiming for does not help :s-smilie:


That is great thanks. It's solved it! :smile:
Reply 5
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the bear
20
Original post by LauraDG
That is great thanks. It's solved it! :smile:


you do not have to use latex for powers....

2[sup]k[/sup}

works fine... just replace the curly bracket with a square one :wink:
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by LauraDG
That is great thanks. It's solved it! :smile:


:biggrin:

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