Consider the series n=1 3 n+ 3 Let Sn be the n-th partial sum; that is, Sn - n 3 i=1 i + 3

Q: Find the sum of the series: 7=0 (-1)"36"+2 (2n + 1)!

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Q: 2n + 3" ind the sum of the series:

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Q: b) Find the sum of the series 10" -1 Σ n =1 11"+1 The sum = c) Find the exact partial sum S, of the…

A: We know that in a geometric series, the sum of infinite terms = a1-r and the sum of first n terms =…

Q: 4 Σ Find the sum of the series (2n - 1)(2n + 1) n= 1

A: The explanation of the solution as follows:

Q: Find the sum of the following series 10 (k+ 2) (k-3) (1 - 2k). k=6

A: The objective of the question is to find the sum of the series 10 * (k + 2) * (k - 3) * (1 - 2k) for…

Q: Find the sum of the series

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Q: 7 + 10 + 13 + . + 40 i-

A: 7+10+13+....+40  First term⇒ a1=7, Last term⇒ an=40common differnce⇒10-7=3 Last term=First…

Q: n E(-2²+2i+2n² + 2n) = i=4

A: Here we will use the concept of summation of n numbers. We will use the formula of sum of n^2 and n…

Q: What is the sum of the series?

A:

Q: can you help me with B,

A: Solving for the part B,

Q: Use sigma notation to write an infinite series whose first foursuccessive partial sums are 10, 20,…

A: As we know, A partial sum is the sum of part of the sequence. n Partial sum is the sum of the first…

Q: Find the sum of the series: n=0 n-3 (-1)^(2/5) " 2n + 1

A: We know     (1 + x)-1 =  ∑n=0∞(−1)nxn   for |x| < 1 => (1 + x2)-1 =  ∑n=0∞(−1)nx2n   So,…

Q: If sine = a, then csce = 01 O a O 1/a

A: cscθ is inverse of sinθ

Q: The sum of the geometric series Two (n=2 4n+1 n=0 5n-2 is

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Q: Find the sum of the series :

A: ∑k=0∞ 1k+1k+3 Apply Series Comparison Test ∑k=0∞ 1k+1k+3≤∑k=0∞ 1k2 ∑k=1∞ 1k2 use p-Series Test If…

Q: Find the sum of each series

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Q: 8] n=1 1 5n2 + 5n

A: To Find: The sum ∑n=1∞15n2+5n

Q: Suppose young Gauss had been asked to compute the sum 6 + 12 + 18 + 24 + ... + 1,494 + 1,500 1. How…

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Q: Calculate the sum (not the partial sums) of the first 10 milion terms in the harmonic series 1 1 + 2…

A: n=1:10^6;L = length(n);sum double = 0;for i=1:Lsum double=sumdouble+double(1/i);%summing in double…