Question

Let C(n) be the constant term in the expansion of
(x + 9)^n.
Prove by induction that
C(n) = 9^n
for all n is in N.
(Induction on n.) The constant term of
(x + 9)^1 is ___ =9 ___
Suppose as inductive hypothesis that the constant term of (x + 9)^k − 1 is____ for some k > 1.
Then (x + 9)^k-1 = (x + 9)^k − 1 · (_____), so its constant term is ____ · 9=_____, as required.

Answer 1

@terenzreignz

Answer 2

@Spacelimbus

Answer 3

can i try?

Answer 4

lol ofcourse i need all the help i can get..

Answer 5

Basic step: C(1) true, it means the constant term in \[(x + 9)^1 ~is~ 9^1\]
Induction step :
Hypothesis step: Assume that C (k-1) is true, it means the constant term in \[(x+9)^{k-1}~is~9^{k-1}\]
What we need to prove that C(k) is true, it means the constant term in \[(x+9)^k ~is~ 9^k\]from hypothesis step. \[(x + 9)^{k-1}*(x+9)\] has the constant term is \[9^{k-1} * 9 = 9^k\] and from the left hand side it is\[ (x+9)^k\]
it implies that 9^k is constant term of (x+9)^k
Proof stop here.
Does it make sense to you?

Answer 6

constant term in expansion polynomial is the last term of it

Answer 7

Good explanation!! Thanks @Loser66 !!

Answer 8

this is more explanation for the last step.

Answer 9

thank u sir

Answer 10

yw