Question

$1+2+3+...............+n$

• A. $=\frac{1}{8}(2n+1{)}^2$
• B. $<\frac{1}{8}(2n+1{)}^2$
• C. $>\frac{1}{8}(2n+1{)}^2$
• D. None of these

Answer 1

The correct option is B

$<\frac{1}{8}(2n+1{)}^2$

For $n=1$
$\frac{1}{8}(2n+1{)}^2=\frac{9}{8}$
$\frac{9}{8}>1$
Let $P\left(n\right)=1+2+3+..............+n<\frac{1}{8}(2n+1{)}^2$
$P\left(1\right)$ is true.
Let $P\left(k\right)$ be true
$\Rightarrow 1+2+3+.........+k<\frac{1}{8}(2k+1{)}^2$
$P(k+1)=1+2+3+.........+k+k+1$
$P(k+1)<\frac{1}{8}(2k+1{)}^2+k+1$
$=\frac{1}{8}\left(\right(2k+1{)}^2+8(k+1))$
$=\frac{1}{8}(4k^2+1+4k+8k+8)$
$=\frac{1}{8}(4k^2+12k+9)$
$P(k+1)<\frac{1}{8}(2k+3{)}^2=\frac{1}{8}(2(k+1)+1{)}^2$
$P(k+1)$ is true
$\therefore P(k+1)$ is true $\forall n\in N$