(a) (1)/(1.3 .5)+(1)/(3.5 .7)+…+(1)/(41.43 .45) is / are equal to (p) 68276 (b) 1^3-2^3+3^3-5^3+

(a) (1)/(1.3 .5)+(1)/(3.5 .7)+…+(1)/(41.43 .45) is / are...

Key Concepts

Telescoping Series

A telescoping series is one in which most terms cancel out sequentially when the series is written in expanded form. This method is particularly useful for evaluating complicated sums by decomposing each term into a difference of two successive terms. The cancellation leads to only a few terms remaining, greatly simplifying the evaluation of the series.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to sum or integrate. When dealing with sums that involve products in the denominator, decomposing them into partial fractions often reveals a telescoping structure or known series, facilitating the calculation of the overall sum.

Summation Formulas

Summation formulas provide closed?form expressions for the sums of sequences such as the first n natural numbers, their squares, cubes, and other powers. These formulas are foundational in simplifying and evaluating series expressions. They are especially useful when a series involves well?known sequences, allowing substitution of the summation formula rather than performing term?by?term addition.

Alternating Series

An alternating series is a series in which successive terms have opposite signs. Evaluating such series often involves recognizing patterns of cancellation or grouping terms strategically. Alternate signs can affect convergence properties and are treated with specific summation techniques to account for the sign alternation.

Products of Consecutive Integers

Products of consecutive integers, such as those formed by multiplying four consecutive numbers, often appear in series and combinatorial identities. Recognizing and simplifying these products can be achieved through patterns or by expressing them in terms of factorials or polynomial forms. This insight simplifies the summation of sequences involving such products, connecting them to known combinatorial expressions and identities.

Transcript

00:01 Well, for the ai part, you find the term, first you find a tn basically, tn is given as 1 over 2n negative 1, then 2n positive 1, then positive 1, then positive 3.

00:18 So, we take now vn equals, the method is 3n, 1 over 2n plus 1, 2n plus 3.

00:27 And then we take v .n negative 1.

00:29 There will be 1 over 2n negative 1, and would be 2n, n, 3.

00:34 Positive 1.

00:35 Next we do vn negative 1 negative vn.

00:38 That's coming out to be 1 over 2n negative 1 2n positive 1 negative 1 over 2 n positive 1 2n positive 3.

00:51 So if we just i mean work on this we take the lcm that's coming out to be 2n negative 1 of 4 2n negative 1 2n negative 1 and 2n positive 3 so it might be 2n positive 3, negative will be 2n positive 1.

01:13 So that is having out to be we have basically 4 times tn.

01:20 And from here we find out tn that equals 1 over 4 vn negative 1 negative vn.

01:28 Now summation is going from we have the last term.

01:33 You look at the last term.

01:35 That is coming out to be 41.

01:37 Right so we start with one and going till 2n negative 1 so we go till 21 so you go like this as then equals we have 1 over 4 then summation and equals 1 to 21 and we get it as vn negative 1 negative vn just work on this we get 1 over 4 then it would be v0 negative v1 positive v1 negative v2 positive and so on negative v2 negative v21 this port here so it's coming out to be 1 over 4 we have then v0 this cancels out ordinate term v0 negative v21 that is coming out to be we have one over four times now we just look at vn the vn we had four it's given as 1 over 2n plus 1 2n plus 3 so v 0 would be 1 over 3 negative v 21 that's coming up to be 1 over 43 times 21 times 2 that is 42 plus 3 45 45 this we got here and to simplify this that's coming out to be 16 1 9 35 so a what is 112 2 s next for the v part.

03:16 So the v part we can write 1 cube negative 2 cube plus 3 cube negative and so on 51 cube equal 1 qib positive 2 cube positive 3 cube positive 51 cube negative 2 times 2 cube positive 4 cube and so on plus 50 this way we can rearrange with nothing we just add here 2 cube and just 2 negative 2 times 2 cube here and this and so on.

03:45 So now we just work on this.

03:49 This will give equal to 1 cube plus 2 cube plus 3 cube and so on plus 51 cube negative 16 times 1 cube plus 2 cube plus and so on plus 25 cube.

04:04 Here we got some of cube of numbers here.

04:06 It is still 25 is still 51...

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