Question

prove that by MI 1.2+2.3+3.4+_____to n terms=n/3(n+1)(n+2)

Answer 1

Question:-

Prove by Principle of Mathematical Induction that, $\displaystyle\sf {P(n):1\cdot2+2\cdot3+3\cdot4+\ \dots\ +n(n+1)=\dfrac {n(n+1)(n+2)}{3}\ \forall n\in\mathbb {N}}$

Solution:-

Let,

$\displaystyle\longrightarrow\sf {P(n):1\cdot2+2\cdot3+3\cdot4+\ \dots\ +n(n+1)=\dfrac {n(n+1)(n+2)}{3}\ \forall n\in\mathbb {N}}$

Consider P(1).

$\displaystyle\longrightarrow\sf {LHS: 1\cdot2=2}$

$\displaystyle\longrightarrow\sf {RHS: \dfrac {1\cdot2\cdot3}{3}=2}$

P(1) is true. So assume P(k) is true.

$\displaystyle\longrightarrow\sf {P(k):1\cdot2+2\cdot3+3\cdot4+\ \dots\ +k(k+1)=\dfrac {k(k+1)(k+2)}{3}\ \forall k\in\mathbb {N}}$

Consider P(k + 1).

$\displaystyle\longrightarrow\sf {P(k+1):1\cdot2+2\cdot3+3\cdot4+\ \dots\ +(k+1)(k+2)=\dfrac {(k+1)(k+2)(k+3)}{3}\ \forall k\in\mathbb {N}}$

Let's check whether it's true.

$\displaystyle\longrightarrow\sf {LHS}$

$\displaystyle\longrightarrow\sf {1\cdot2+2\cdot3+3\cdot4+\ \dots\ +k(k+1)+(k+1)(k+2)}$

$\displaystyle\longrightarrow\sf {\dfrac {k(k+1)(k+2)}{3}+(k+1)(k+2)}$

$\displaystyle\longrightarrow\sf {(k+1)(k+2)\left [\dfrac {k}{3}+1\right]}$

$\displaystyle\longrightarrow\sf {\dfrac {(k+1)(k+2)(k+3)}{3}}$

$\displaystyle\longrightarrow\sf {RHS}$

Therefore P(k + 1) is true whenever P(k) is true.

Hence P(n) is true $\displaystyle\sf {\forall n\in\mathbb{N}.}$