Hi,
Can someone pls help to solve the below question asap.
Prove the statement is true using the mathematical induction.
2 + 4 + 6 +..... + 2n = n (n +1)
Best Regards
Lalit Chugh[/B]
Hello,Originally Posted by lalitchugh
I presume, that you know how to do a proof by induction.
A(1): n = 1. That means your equation becomes: 2 = 1*(1+1) which is true.
A(2): Let 2+4+6+...+2n = n(n+1) be true.
A(3): Proof that 2+4+6+...+2n+2(n+1) =n(n+1)+2(n+1). Factor the RHS
2+4+6+...+2n+2(n+1) =(n+1)(n+2)
2+4+6+...+2n+2(n+1) =(n+1)(n+1+1)
That means you've changed successfully the n's at the RHS into (n+1)'s. So this equation is true for all $\displaystyle n\in \mathbb{N}$
Greetings
EB
Hello, Lalit!
Are you a bit shaky on Induction?
This one is fairly simple . . .
Prove the statement is true using the mathematical induction.
. . . $\displaystyle 2 + 4 + 6 + \hdots + 2n \;= \;n(n +1)$
Verify $\displaystyle S(1):\;\;2 \:= \:1(1 + 1) \:= \:2$ . . . True!
Assume $\displaystyle S(k):\;\; 2 + 4 + 6 + \hdots + 2k \;= \; k(k + 1)$
Add $\displaystyle 2(k + 1)$ to both sides:
. . . $\displaystyle 2 + 4 + 6 + \hdots + 2k + 2(k + 1) \;= \;k(k + 1) + 2(k+1)$
. . . $\displaystyle 2 + 4 + 6 + \hdots + 2(k+1) \;= \;k^2 + k = 2k + 2$
. . . $\displaystyle 2 + 4 + 6 + \hdots + 2(k+1) \;=\;(k+1)(k+2)$ . . . This is $\displaystyle S(k+1)\,!$