Can someone tell me how the correct answer for this question is deduced?

This is correct:
\(\sum\limits_{j = 1}^n {(5{j^2} + 3j - 1)} = 5\sum\limits_{j = 1}^n {{j^2}} + 3\sum\limits_{j = 1}^n j - \sum\limits_{j = 1}^n 1 \\=\dfrac{5(n+1)(2n-1)}{6}+\dfrac{3n(n+1)}{2}-n\)
 
Last edited:
pka said:
This is correct:
\(\sum\limits_{j = 1}^n {(5{j^2} + 3j - 1)} = 5\sum\limits_{j = 1}^n {{j^2}} + 3\sum\limits_{j = 1}^n j - \sum\limits_{j = 1}^n 1 \)
Click to expand...

Hi, thanks for the reply. Could you please tell me how you came to that answer? I'm really struggling to understand. Please let me know
 
It is a bad set of responses.

The first response is technically correct (except perhaps for formatting). You always can, and frequently should, break down such a sum into individual sums

[MATH]\sum_{i=1}^n (5i^2 + 3i - 1) \equiv 5\sum_{i=1}^n i^2 + 3 \sum_{i=1}^n i - \sum_{i=1}^n 1.[/MATH]
although I prefer to put parentheses around the individual sums. However, it is a useless answer.

The second answer is just plain wrong.

The third answer is correct and useful. So that is the one they want.
 
bushra1175 said:
[MATH]\sum\limits_{j = 1}^n {(5{j^2} + 3j - 1)} = 5\sum\limits_{j = 1}^n {{j^2}} + 3\sum\limits_{j = 1}^n j - \sum\limits_{j = 1}^n 1[/MATH]Hi, thanks for the reply. Could you please tell me how you came to that answer? I'm really struggling to understand. Please let me know
Click to expand...
This is the same as if I said, [MATH](5x^2+3x-1)+(5y^2+3y-1) = 5(x^2+y^2)+3(x+y)-(1+1)[/MATH]. Can you see why that is true? There some of each of the associative, commutative, and distributive properties.
 
Why is the first one correct.

[MATH]\sum_{i=1}^2 (5i^2 + 3i - 1) =\\ (5 * 1^2 + 3 * 1 - 1) + (5 * 2^2 + 3 * 2 - 1) = \\ (5 * 1^2 + 5 * 2^2) + (3 * 1 + 3 *2) + (-1) + (-1) =\\ 5(1^2 + 2^2) + 3(1 + 2) - (1 + 1) =\\ 5\left ( \displaystyle \sum_{i=1}^2 i^2 \right ) +3 \left ( \displaystyle \sum_{i=1}^2 i \right ) - \left (\displaystyle \sum_{i=1}^2 1 \right ).[/MATH]You are just rearranging the terms of the sum and then factoring out common multiples.

But suppose we use the formula in the third line.

[MATH]\sum_{i=1}^2 (5i^2 + 3i - 1) =\\ (5 * 1^2 + 3 * 1 - 1) + (5 * 2^2 + 3 * 2 - 1) = \\ 5 * 1 + 3 * 1 - 1 + 5 * 4 + 3 * 2 - 1 = \\ 5 + 3 - 1 + 20 + 6 - 1 = 7 + 25 = 32 =\\ 25 + 9 - 2 =\\ 5 * 5 + 3 * 3 - 2 =\\ 5 * \dfrac{2(2 + 1) (2 * 2 + 1)}{6} + 3 * \dfrac{2(2 + 1)}{2} - 2.[/MATH]The formula does not save you any time when n = 2. But suppose n = 23?

Anyway, when they give you two correct answers and ask you to pick one, it's stupid.

EDIT: Wait a minute. I am the stupid one. They say "which represent" instead of "which represents" and thereby indicate that more than one answer is correct. It's also an English test!
 
Last edited:
JeffM said:
Why is the first one correct.

[MATH]\sum_{i=1}^2 (5i^2 + 3i - 1) =\\ (5 * 1^2 + 3 * 1 - 1) + (5 * 2^2 + 3 * 2 - 1) = \\ (5 * 1^2 + 5 * 2^2) + (3 * 1 + 3 *2) + (-1) + (-1) =\\ 5(1^2 + 2^2) + 3(1 + 2) - (1 + 1) =\\ 5\left ( \displaystyle \sum_{i=1}^2 i^2 \right ) +3 \left ( \displaystyle \sum_{i=1}^2 i \right ) - \left (\displaystyle \sum_{i=1}^2 1 \right ).[/MATH]You are just rearranging the terms of the sum and then factoring out common multiples.

But suppose we use the formula in the third line.

[MATH]\sum_{i=1}^2 (5i^2 + 3i - 1) =\\ (5 * 1^2 + 3 * 1 - 1) + (5 * 2^2 + 3 * 2 - 1) = \\ 5 * 1 + 3 * 1 - 1 + 5 * 4 + 3 * 2 - 1 = \\ 5 + 3 - 1 + 20 + 6 - 1 = 7 + 25 = 32 =\\ 25 + 9 - 2 =\\ 5 * 5 + 3 * 3 - 2 =\\ 5 * \dfrac{2(2 + 1) (2 * 2 + 1)}{6} + 3 * \dfrac{2(2 + 1)}{2} - 2.[/MATH]The formula does not save you any time when n = 2. But suppose n = 23?

Anyway, when they give you two correct answers and ask you to pick one, it's stupid.

EDIT: Wait a minute. I am the stupid one. They say "which represent" instead of "which represents" and thereby indicate that more than one answer is correct. It's also an English test!
Click to expand...

Thank you so much. I can clearly see now how the first option equates to the original sum notation. However, I am confused about how you came to deduce that n = 2 for the third line. Was it by trial and error?

I can see in your breakdown that the highlighted part is just a simplification of the original sum notation. but I don't see how you plugged it in the formula with n being 2 below
1592356268219.png
 
1592356268219.png

I am not sure what you are asking. n was just replaced with 2.
 
bushra1175 said:
Thank you so much. I can clearly see now how the first option equates to the original sum notation. However, I am confused about how you came to deduce that n = 2 for the third line. Was it by trial and error?

I can see in your breakdown that the highlighted part is just a simplification of the original sum notation. but I don't see how you plugged it in the formula with n being 2 below
View attachment 19781
Click to expand...
He gave an example (as I did, also, in post #5) showing what the sum means in the case n=2 (that is, just two terms). So that was chosen from the start.

I myself would not have written in the way he did; I would write his last three lines in reverse order, showing how the formula in the third option would be evaluated, resulting in the fact that both the summation and the formula give the same result, 32. This doesn't prove the formula, but is supporting evidence that it is correct.
 
Jomo said:
1592356268219.png

I am not sure what you are asking. n was just replaced with 2.
Click to expand...
Sorry for the confusion. I was asking why he replaced n with 2 and not any other number to show that the formula is correct, but I can see now that its because its 2 over here:
1592365638419.png


Dr.Peterson said:
He gave an example (as I did, also, in post #5) showing what the sum means in the case n=2 (that is, just two terms). So that was chosen from the start.
Click to expand...

Thank you Dr. Peterson for making me realise this
 
You must log in or register to reply here.
Top