induction (1 Viewer)

[twistedrebel]

I got 2 quick induction questions

1. 1/1.2 + 1/2.3 + 1/3.4 .....+1/n(n+1) = 1 - 1/n+1


Another question is
2. n(sigma notation) 1 n^3= n^2(n+1)^2/4

 

[ninetypercent]

For n =1,
LHS = 1/(1)(2) = 1/2
RHS = 1 - 1/(1+1) = 1/2
LHS = RHS
therefore, true for n = 1

assume true for n = k

1/1.2 + 1/2.3 + 1/3.4 .....+1/k(k+1) = 1 - 1/k+1

rtp true for n = k + 1

1/1.2 + 1/2.3 + 1/3.4 .....+1/k(k+1) + 1/(k+1)(k+2) = 1 - 1/k+2

LHS = Sk + Tk + 1
= 1 - 1/k+1 + 1/(k+1)(k+2)


the statement is true for n = k + 1, if it is true for n = k

therefore, [insert conclusion]

 

[ninetypercent]

n^3= n^2(n+1)^2/4

For n =1
LHS = 1^3 = 1
RHS = 1(1+1)^2/4 = 4/4 = 1
LHS = RHS
therefore true for n =1

assume true for n = k
(sigma) k^3= k^2(k+1)^2/4

RTP true for n = k + 1




the statement is true for n = k + 1, if it is true for n = k

therefore, [insert conclusion]

 

[fullonoob]

damn the rabbit is travelling too fast, last time i saw it was 8 months left ><

 

[spyborg]

lol replace the rabbit with a snail

 

[ninetypercent]

still half a year left to go

 

[hello-there]

Help needed plz
Induction Q.
Prove n^2-2n+5>0

 

[ninetypercent]

when n = 1
1 -2 + 5 = 4 > 0
therefore, true for n =1

assume true for n = k
k^2 - 2k + 5 > 0

RTP true for n = k + 1
(k+1)^2 - 2(k+1) + 5 > 0

LHS = (k+1)^2 - 2(k+1) + 5
= k^2 + 2k + 1 - 2k -2 + 5
= (k^2 - 2k + 5) + 2k + 1
> 2k + 1 (from assumption)
> 0 since k> 0
= RHS

the statement is true for n = k + 1, if it is true for n = k

therefore, [insert conclusion]

 

[hello-there]

when n = 1
1 -2 + 5 = 4 > 0
therefore, true for n =1

assume true for n = k
k^2 - 2k + 5 > 0

RTP true for n = k + 1
(k+1)^2 - 2(k+1) + 5 > 0

LHS = (k+1)^2 - 2(k+1) + 5
= k^2 + 2k + 1 - 2k -2 + 5
= (k^2 - 2k + 5) + 2k + 1
> 2k + 1 (from assumption)
> 0 since k> 0
= RHS

the statement is true for n = k + 1, if it is true for n = k

therefore, [insert conclusion]

umm can u explain the underlined part for me and theres a mistake with the bolded stuff

much appreciated man