Question

4. Using principle of mathematical induetlon, prove that 1+2 +3+ n =n(n+1) /2
where n is a natural number,​

Answer 1

Given :

$\displaystyle { 1+2+3+...n=\frac{n(n+1)}{2},n\in N.}$

To Find :

• By using principal of mathematical induction prove the given equation.

Solution :

Let the given statement $P(n)$ be defined as

$\displaystyle {\sf 1+2+3+...n=\frac{n(n+1)}{2},n\in N.}$

Step 1 :-

$\sf Put \: n=1 \\\\\sf LHS \: P(1)=1\\\\\sf RHS \:P(1)=\dfrac{1(1+1)}{2}=1\\\\\sf LHS=RHS \:for\:n=1\\\\\sf\therefore \:P(1)\:is\:true$

Step 2 :-

Let us assume that the statement is true for $(n=k).$

$\sf i . e ., P(k)\:is\:true$

$\displaystyle {\sf 1+2+3+...k=\frac{k(k+1)}{2}}$ is true.

Step 3 :-

$\sf P(k+1)=1+2+3+ ... +k+(k+1)$

$\sf \longrightarrow P(k)+k+1\\\\\longrightarrow \sf\dfrac{k(k+1)}{2}+k+1\\\\\longrightarrow \sf\dfrac{k(k+1)+2(k+1)}{2}\\\\\longrightarrow\sf \dfrac{(k+1)(k+2)}{2}\\\\\therefore\sf P(k+1)\:is\:true$

Thus if $\sf P(k)$ is true, then $\sf P(k+1)$ is also true.

$\sf \therefore P(n) \:is \: true \:for\:all\:n\in N$

Hence $\displaystyle { \sf1+2+3+...n=\frac{n(n+1)}{2},n\in N.}$

Answer 2

To ProvE :-

• By the principle of mathematical induction , to Prove that ,

$\sf 1 + 2 + 3 + 4 + .. ... + n =\dfrac{n(n+1)}{2}$

ProoF :-

The process of mathematical induction is a indirect method which helps us to prove some formulae which cannot be proved by direct methods .

$\red{\bigstar}\underline{\textsf{ Principal of Mathematical Induction :- }}$

$\textsf{ If P(n) is a statement such that , }$

• P(n) is true for n = 1
• P(n) is true for n = k + 1 , when it's also true for n = k , where $\sf k \in \mathbb{N}$

Then we can say that by the principal of mathematical induction , the statement is true for all Real Numbers .

$\rule{200}2$

$\sf \to Let \ \blue{P(n)} : 1 + 2 + 3 + 4 + ... .. + n =\dfrac{n(n+1)}{2}$

$\red{\bigstar} \textsf{ \textbf{Step 1} :\underline{ Put n = 1 :-}}$

Then , LHS = 1 , and

RHS = $\sf \dfrac{1+1}{2}(1)=\dfrac{2}{2}=\orange{1} .$

$\textsf{ \therefore \textbf{ LHS}=\textbf{RHS} . }$

$\dashrightarrow\textsf{\purple{So P(n) is true for n = \textbf{1}}.}$

$\rule{200}2$

$\red{\bigstar} \textsf{ \textbf{Step 2 } : \underline{Assume that P(n) is true for n = k . }}$

$\to \sf 1 + 2 + 3 + 4 + ... . .. + k = \orange{\dfrac{k(k+1)}{2}}$

$\to \sf 1 + 2 + 3 + 4 + .. ... . + k + (k + 1 ) = \dfrac{k(k+1)}{2}+(k+1) \qquad \bigg\lgroup \red{\tt Adding\ (k+1) \ both \ sides } \bigg\rgroup$

• So that :-

$\sf\to (k+1)\bigg( \dfrac{k}{2}+1\bigg) \\\\\sf\to (k+1)\bigg( \dfrac{k+2}{2}\bigg) \\\\\sf\to \dfrac{(k+1)(k+2)}{2} \\\\\sf\to \dfrac{(k+1)(\red{k+1}+1) }{2}$

$\sf\dashrightarrow \purple{ P(n) \ is \ also \ true \ for \ n = \ k + 1 }$

Therefore by the principal of Mathematical Induction we can say that , P(n) is true for all natural numbers .

$\rule{200}2$

$\large\boxed{\pink{\sf\displaystyle\sum^{\tt n}_{\tt{k=1}}=\dfrac{\sf n(n+1)}{\sf 2} \ \sf for \ all \ n \ \in \mathbb{N} }}.$

$\rule{200}2$