Induction Inequality Proof: 3^n is greater than or equal to 2n + 1

[@Gizmare0256]

Thank the lord this video found me, you saved me hours of pain yet again!

[@Voy2k23]

i am so confused as to how and why inequality sign changed mid calculations.

[@Grand-pq]

since 3^k is already >= , it would make sense that 3^(k+1)> . If K is greater then, then k+1 is definitely greater. While if K is equal to, then k+1 would also be greater, if this is what you ment.

[@autumnmiloser2709]

can you show 3(2k + 2 + 1) by adding values to left side (this should be ok because it's greater than) and then simply say 3(2k + 2 + 1) is greater than 2(k + 1) + 1 because our domain is positive integers and the same equation * 3 is greater than the one that's not

[@preranabhoi3325]

Thank you so much Sir . This video is really very helpful for me.

[@dccarajay]

Sir, I am literally thankful of this channel and your existence. Thank you!

[@lasgill]

😂

[@maths3630]

You must used (1+x)^n>=1+nx n€IN and x€IR+

[@niyya9932]

VERY WELL SIMPLIFIED. THANK YOU SIR

[@haydeehernandez3763]

inequality n² ≤ 9 = 0,1,2 and 3 I know is -1, -2 and -3 but can someone explain it why or show me step by step

[@jianfenglei571]

can the base case be 0 that 3^0 = 1?

[@Xuri-u9j]

Natural numbers start from 1 not zero

[@Sellipsis]

6:42 the explanation for this makes no sense, you kept the multiplication of 3*1 in 2k+3 but you're free to throw away 3x2k? Where did it go? Why is one okay without the other?

[@phoemi]

it's not an equation, he's showing that (3 * 2k + 3) > (2k + 3)

[@yewmin5769]

For example, 3a > a when a is Z+.

[@petzClaver109]

thank you, i didn't get that part

[@jeffreychan4442]

ur thumbnail is different from the proof u do btw lool

[@TheMathSorcerer]

omg! I will fix it, thank you!!!!!!!!!!!!!!!!!!!!!