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1 + 2 + 3 + ... + n =

Question:

$1+2+3+\ldots+n=\frac{n(n+1)}{2}$ i.e. the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$

Solution:

Let P(n) be the given statement.

Now,

$\mathrm{P}(n)=1+2+3+\ldots+n=\frac{n(n+1)}{2}$

Step 1: 

$P(1)=1=\frac{1(1+1)}{2}=1$

Hence, $P(1)$ is true.  

Step 2: 

Let $P(m)$ be true.

Then,

$1+2+3+\ldots+m=\frac{m(m+1)}{2}$

We shall now prove that $P(m+1)$ is true.

i. e.,

$1+2+\ldots+(m+1)=\frac{(m+1)(m+2)}{2}$

Now,

$1+2+\ldots+m=\frac{m(m+1)}{2}$

$\Rightarrow 1+2+\ldots m+m+1=\frac{m(m+1)}{2}+m+1 \quad$ [Adding $(m+1)$ to both sides]

$=\frac{m^{2}+m+2 m+2}{2}$

$=\frac{(m+1)(m+2)}{2}$

Hence, $P(m+1)$ is true.

By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathrm{N}$.

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