1+2+3+....n lt; 18(2n+1)^2 ⇒ n (n+1)2 lt; 18 (2n+1)^2 4n^2+4n lt; 4n^2+1+4n 0 lt; 1 Hence proved ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Summation by Sigma Method

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

Question

$1 + 2 + 3 + . . . . + n < \frac{1}{8} (2 n + 1)^{2}$ .

Open in App

Solution

$1 + 2 + 3 + . . . . n < \frac{1}{8} (2 n + 1)^{2}$
$\Rightarrow \frac{n (n + 1)}{2} < \frac{1}{8} (2 n + 1)^{2}$
$4 n^{2} + 4 n < 4 n^{2} + 1 + 4 n$
$0 < 1$
Hence proved

Suggest Corrections

Similar questions

1 + 2 + 3 + . . . . . . . . . . . . . . n < \frac{1}{8} (2 n + 1)^{2}

n

2^{4 n - 2} = \frac{1}{8^{2 n - 6}}

1 + 2 + 3 + 4 + . . . . . + n = ?

1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?

1 + 1 + 1 + 1 + . . . . . + n = ?

1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?

\frac{3^{n} - 1}{2}

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + \dots + n

=

Join BYJU'S Learning Program