# Summation

#### helloying

Show that $$\displaystyle \sum (-1)^{r+1} r^2$$ = -n(2n+1)

summation is from r=1 to 2n . sry i dont know how to type

Thx Show that $$\displaystyle \sum (-1)^{r+1} r^2$$ = -n(2n+1)

summation is from r=1 to 2n . sry i dont know how to type

Thx hi

\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate

$$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$$

$$\displaystyle =1-4+9-16+25-36+...$$

$$\displaystyle =1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$$

$$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n-1)$$

$$\displaystyle =-3-7-...-(4n-1)$$

which is an AP with first term -3 , last term -(4n-1)

$$\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$$

simplify this and you are done.

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• helloying

#### oldguynewstudent

hi

\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate

$$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$$

$$\displaystyle =1-4+9-16+25-36+...$$

$$\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$$

$$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$$

$$\displaystyle =-3-4-...-(4n-1)$$

which is an AP with first term -3 , last term -(4n-1)

$$\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$$

simplify this and you are done.

How did you get $$\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$$?

When I do this I come up with $$\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$$, because $$\displaystyle (-1)^{2n+1} {2n}^2$$ = $$\displaystyle -{2n}^2$$.

#### helloying

How did you get $$\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$$?

When I do this I come up with $$\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$$, because $$\displaystyle (-1)^{2n+1} {2n}^2$$ = $$\displaystyle -{2n}^2$$.
yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$$

it shd be -3-7 and not -3-4

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yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$$