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Find the first four partial sums, s1,s2,s3,s4, and the nth partial sum of the squence an=log(nn+1).

Sagot :

The first four partial sums of the sequence are S₁ = 0.3010, S₂ = 0.9999, S₃ = 2.447, S₄ = 4.8569.

In this question,

A sequence is a set of things (usually numbers) that are in order. A partial sum is the sum of part of the sequence.

The sequence is [tex]a_{n} =log(n^{n}+1 )[/tex]

The first four partial sum S₁, S₂, S₃, S₄ can be calculated by substituting n = 1,2,3,4 in the sequence.

S₁ can be calculated as

S₁ = a₁

⇒ [tex]a_{1} =log(1^{1}+1 )[/tex]

⇒ [tex]a_{1} =log(1+1 )[/tex]

⇒ [tex]a_{1} =log(2 )[/tex]

⇒ [tex]a_{1} =0.3010[/tex]

Now, S₁ = 0.3010

S₂ can be calculated as

S₂ = a₁ + a₂

⇒ [tex]a_{2} =log(2^{2}+1 )[/tex]

⇒ [tex]a_{2} =log(4+1 )[/tex]

⇒ [tex]a_{2} =log(5 )[/tex]

⇒ [tex]a_{2} =0.6989[/tex]

Now, S₂ = 0.3010 + 0.6989

⇒ S₂ = 0.9999

S₃ can be calculated as

S₃ = a₁ + a₂ + a₃

⇒ [tex]a_{3} =log(3^{3}+1 )[/tex]

⇒ [tex]a_{3} =log(27+1 )[/tex]

⇒ [tex]a_{3} =log(28)[/tex]

⇒ [tex]a_{3} =1.4471[/tex]

Now, S₃ = 0.3010 + 0.6989 + 1.4471

⇒ S₃ = 2.447

S₄ can be calculated as

S₄ = a₁ + a₂ + a₃ + a₄

⇒ [tex]a_{4} =log(4^{4}+1 )[/tex]

⇒ [tex]a_{4} =log(256+1 )[/tex]

⇒ [tex]a_{4} =log(257 )[/tex]

⇒ [tex]a_{4} =2.4099[/tex]

Now, S₄ = 0.3010 + 0.6989 + 1.4471 + 2.4099

⇒ S₄ = 4.8569

Hence we can conclude that the first four partial sums of the sequence are S₁ = 0.3010, S₂ = 0.9999, S₃ = 2.447, S₄ = 4.8569.

Learn more about partial sum of the sequence here

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