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phospho
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I don't understand how the part in yellow can give you U_n, I just don't see how taking the two summations away from each other would give U_n, could anyone explain it please
micromass said:Write it out for n=4. You'll see immediately what happens.
I think micromass meant this:phospho said:write what out? I've substituted n = 4 and get 20 if I use what they have used for part b...
eumyang said:Think about it.
[itex]\Sigma^{n}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n[/itex], and
[itex]\Sigma^{n - 1}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1}[/itex].
So what happens when you subtract the two summations:
[itex]\left( U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n \right) - \left( U_1 + U_2 + U_3 + ... + U_{n - 1} \right)[/itex]?
EDIT: Beaten to it.
[itex]\Sigma^{n}_{r = 1} U_r = U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n = n^2 + 4n[/itex]. That was given in the problem. Notice the substitution that was made in the step after the highlighted step.phospho said:I see, but what is the "n^2 + 4n", is that a general term or..?
U_n refers to the nth term in a series, where the value of n can vary depending on the specific problem. It is a variable that represents a specific element in the series.
The use of two summations allows for more complex series questions to be solved, as it takes into account both the general term of the series and the specific term being asked for. It can also provide a more efficient way of solving certain series problems.
The value of n can be determined by looking at the specific element that is being asked for in the series. For example, if the question asks for the 10th term in the series, then n would be equal to 10.
The general formula is U_n = A + (n-1)d, where A is the first term of the series, n is the term being asked for, and d is the common difference between each term in the series. This formula can be used for arithmetic series, while other types of series may have different formulas.
No, this method is specifically for solving series questions that involve two summations and have a constant difference between each term. Other types of series may require different methods or formulas for solving them.