sum from n=0 to infinity of ((3+2n))/(2^n)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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∑_n=0^∞(3+2n)2ⁿ

Solution

converges

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∑_n=0^∞3+2n2ⁿ

Expand 3+2n2ⁿ : 32ⁿ +2¹⁻ⁿn

=∑_n=0^∞32ⁿ +2¹⁻ⁿn

Apply the Sum Rule: ∑a_n+b_n=∑a_n+∑b_n

Series Sum/Difference Rule description:

If ∑a_n and ∑b_n are both convergent, then ∑a_n±b_n=∑a_n±∑b_n

=∑_n=0^∞32ⁿ +∑_n=0^∞2¹⁻ⁿn

∑_n=0^∞32ⁿ =6

∑_n=0^∞2¹⁻ⁿn=converges

=6+converges

=converges

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

sum from n=0 to infinity of ((3+2n))/(2^n)

Solution

Solution steps

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