If P n is the statement -n n 2- n is even- and if P r is true- then P r -1 is true-

Here, P (n) : n^2 + n is even Given, P (r) is true ⇒ r^2 + r is even ⇒ r^2 + r = 2 λ ...........(1) Now, (r + 1)^2 + (r + 1) =r^2 + 2r + 1 + r + 1 = (r^ ...

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Standard XII

Mathematics

Proof by mathematical induction

If P n is the...

Question

If P(n) is the statement "n^2 + n is even", and if P (r) is true, then P (r + 1) is true.

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Solution

Here, P (n) : $n^{2} + n$ is even

Given, P (r) is true

$\Rightarrow r^{2} + r$ is even

$\Rightarrow r^{2} + r = 2 λ$ ...........(1)

Now,

$(r + 1)^{2} + (r + 1)$

$= r^{2} + 2 r + 1 + r + 1$

$= (r^{2} + r) + 2 r + 2$

$= 2 λ + 2 r + 2 [U s i n g e q u a t i o n (1)]$

$= 2 (λ + r + 1)$

$= 2 μ$

$\Rightarrow (r + 1)^{2} + (r + 1)$ is even

$\Rightarrow p (r + 1)$ is true.

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If P(n) is the statement "n^2 + n is even", and if P (r) is true, then P (r + 1) is true.

Here, P (n) : n2+n is even Given, P (r) is true ⇒r2+r is even ⇒r2+r=2λ ...........(1) Now, (r+1)2+(r+1) =r2+2r+1+r+1 =(r2+r)+2r+2 =2λ+2r+2 [Using equation(1)] =2(λ+r+1) =2μ ⇒(r+1)2+(r+1) is even ⇒p(r+1) is true.