Question

prove by mathematica
by mathematical indution that 1+2+_____+n=[m(n+1)÷2]​

Answer 1

Step-by-step explanation:

the explanation is given above

Answer 2

Basic Concept Used :-

Principal of Mathematical Induction :-

The technique involves two steps to prove a statement, as stated below −

• Step 1 (Base step) − It proves that a statement is true for the initial value say n = 1.

• Step 2(Inductive step) − It proves that if the statement is true for the kth iteration (or number k), then it is also true for (k+1)th iteration ( or number k+1).

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: P(n) : 1 + 2 + 3 + - - - + n = \dfrac{ n( n + 1)}{2}$

Step :- 1

$\rm :\longmapsto\:For \: n = 1$

$\rm :\longmapsto\:1 = \dfrac{1(1 + 1)}{2} = \dfrac{2}{2} = 1$

$\bf\implies \:P(n) \: is \: true \: for \: n \: = \: 1$

Step :- 2

$\rm :\longmapsto\:Assume \: that \: P(n) \: is \: true \: for \: n = k \: where \: k \in \: N$

$\rm :\longmapsto\:1 + 2 + 3 + - - - + k = \dfrac{ k( k + 1)}{2} - - (1)$

Step :- 3

$\rm :\longmapsto\:For \: n = k + 1 \: we \: have \: to \: prove \:P(n) \: is \: true$

$\rm :\longmapsto\:1 + 2 + 3 + - - - + k + (k + 1) = \dfrac{ (k + 1)( k + 2)}{2}$

$\bf \: Consider \: LHS$

$\rm :\longmapsto\:1 + 2 + 3 + - - - + k + (k + 1)$

$\rm :\longmapsto\: = \dfrac{k(k + 1)}{2} + (k + 1) \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\: = \: (k + 1)\bigg(\dfrac{k}{2} + 1\bigg)$

$\rm :\longmapsto\: = \: (k + 1)\bigg(\dfrac{k + 2}{2} \bigg)$

$\rm :\longmapsto\: = \: \dfrac{(k + 1)(k + 2)}{2}$

$\rm :\longmapsto\: = \: RHS$

$\bf\implies \:P(n) \: is \: true \: for \: n \: = k + 1$

Hence,

$\rm :\longmapsto\:By \: the \: process \: of \: PMI \:$

$\bf:\longmapsto\:1 + 2 + 3 + - - - + n = \dfrac{ n( n + 1)}{2}$