How can be done by the method of mathematical induction?

Question

We are given that $P(x+1)-P(x)=2x+1$ We also know that $P(0)=1$ We want to prove that $P(2004)=(2004)^2 +1$ Can someone explain how can be solved with mathematical induction? Thank you in advance!

Answer 1

Observe that $P(y)=y^2+1$ satisfies the conditions [Proof below]

So, if $P(x)=x^2+1, P(x+1)=P(x)+2x+1=x^2+2x+1+1=(x+1)^2+1$

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[ Proof:

As $P(x+1)-P(x)=2x+1$

$P(x)$ can be at most Quadratic Polynomial

Let $P(x)=Ax^2+Bx+C$ as $P(0)=1, A\cdot0^2+B\cdot0+C+C\implies C=1$

So, $P(x)=Ax^2+Bx+1, P(x+1)=A(x+1)^2+B(x+1)+1$

So, $2x+1=P(x+1)-P(x)=A(2x+1)+B=2Ax+A+B$

Equating the coefficients of $x,2=2A\implies A=1$

Equating the constants $A+B=1\implies B=0$

]

Answer 2

Proof by induction that $P(k) = k^2 + 1$:

Basis: $k = 0$: $P(0) = 1 = 0^2 + 1$, check.

Induction: Suppose it is valid for $k$:

$$ \begin{align*} P(k + 1) &= P(k) + 2 k + 1 \\ &= (k^2 + 1) + 2 k + 1 \\ &= (k^2 + 2 k + 1) + 1 \\ &= (k + 1)^2 + 1 \end{align*} $$ This checks.

By induction, this is then valid for all $k \ge 0$.