Question #c14ae

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Answer 1 · 2017-05-03T17:32:47

See my answer to your previous question about induction.

Answer 2 · 2017-05-03T17:42:31

Assume true for #n = k#.

#therefore T_k = 2+5+8+...+(3k-1)=1/2k(3k+1) qquad triangle#

And so it follows that:

#T_(k+1) =1/2k(3k+1) color(blue)(+ (3(k+1) - 1))#

# =3/2k^2+1/2 k + 3k+ 3 - 1#

# =3/2k^2+7/2 k + 2#

# =1/2 (3k^2+7 k + 4)#

Long divide by #k+1#

# =1/2 (k+1)(3k + 4)#

# =1/2 (k+1)(3(k+1) + 1)#

So if this is true for #n=k#, it is true for #n=k+1, k+2 , k+3, ...#

Next, set #k = 1# and we see that for #star#, LHS = RHS = 2.

So it is true for #k = 1,2,3,...#, ie #k in NN#