Find the sum of the series

What is the sum of the series: 2 + 12 + 40 + 112 + .....................infinity ?

#Algebra #ArithmeticGeometricProgression

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

According to me the answer should be infinity, since the series is increasing and common ratio is greater than 1. But is my reasoning correct? @Brian Charlesworth

Manish Dash - 6 years ago

Yes, that's correct. The $n$ th term is of the form $a(n) = 2^{n}(2n - 1),$ which goes to $\infty$ as $n \rightarrow \infty,$ and so by the $n$ th term test we conclude that the series diverges.

Just out of curiosity, note that the sum of the first $N$ terms of this series is $(2N - 3)2^{N+1} + 6.$

Brian Charlesworth - 6 years ago

Thank You Sir.

Manish Dash - 6 years ago

The series does not converge.

Nihar Mahajan - 6 years ago

ok so is the answer infinity. @Nihar Mahajan

Manish Dash - 6 years ago

I guess yes.

Nihar Mahajan - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$