Answer:
0.8
Step-by-step explanation:
Answer:
im not sure but i think 66
Step-by-step explanation:
101 = 100 + 1
102 = 100 + 2
103 = 100 + 3
and so on, and
99 = 100 - 1
98 = 100 - 2
97 = 100 - 3
and so on. Then the
-th term of the sum, where
, is
![(-1)^{n-1}(100+n)(100-n)=(-1)^n(n^2-100)](https://tex.z-dn.net/?f=%28-1%29%5E%7Bn-1%7D%28100%2Bn%29%28100-n%29%3D%28-1%29%5En%28n%5E2-100%29)
We want to compute the sum,
![(101\cdot99)-(102\cdot98)+\cdots+(149\cdot51)-(150\cdot50)=\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)](https://tex.z-dn.net/?f=%28101%5Ccdot99%29-%28102%5Ccdot98%29%2B%5Ccdots%2B%28149%5Ccdot51%29-%28150%5Ccdot50%29%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5En%28n%5E2-100%29)
We have
![\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\sum_{n=1}^{50}(-1)^nn^2-100\sum_{n=1}^{50}(-1)^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5En%28n%5E2-100%29%3D%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5Enn%5E2-100%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5En)
but notice that in the last sum, we're just adding the same number of 1s and -1s together, so its value is 0 and
![\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\boxed{\sum_{n=1}^{50}(-1)^nn^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5En%28n%5E2-100%29%3D%5Cboxed%7B%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5Enn%5E2%7D)
In case you're not familiar with the formula for the sum of consecutive squares, we can derive it here. Recall that
![\displaystyle\sum_{n=1}^k1=k](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek1%3Dk)
![\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ekn%3D%5Cfrac%7Bk%28k%2B1%29%7D2)
Notice that
![(n+1)^3-n^3=(n^3+3n^2+3n+1)-n^3=3n^2+3n+1](https://tex.z-dn.net/?f=%28n%2B1%29%5E3-n%5E3%3D%28n%5E3%2B3n%5E2%2B3n%2B1%29-n%5E3%3D3n%5E2%2B3n%2B1)
and that
![\displaystyle\sum_{n=1}^k((n+1)^3-n^3)=(2^3-1^3)+(3^2-2^3)+\cdots+(k^3-(k-1)^3)+((k+1)^3-k^3)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek%28%28n%2B1%29%5E3-n%5E3%29%3D%282%5E3-1%5E3%29%2B%283%5E2-2%5E3%29%2B%5Ccdots%2B%28k%5E3-%28k-1%29%5E3%29%2B%28%28k%2B1%29%5E3-k%5E3%29)
![\implies\displaystyle\sum_{n=1}^k((n+1)^3-n^3)=(k+1)^3-1](https://tex.z-dn.net/?f=%5Cimplies%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek%28%28n%2B1%29%5E3-n%5E3%29%3D%28k%2B1%29%5E3-1)
Then
![(k+1)^3-1=\displaystyle\sum_{n=1}^k(3n^2+3n+1)](https://tex.z-dn.net/?f=%28k%2B1%29%5E3-1%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek%283n%5E2%2B3n%2B1%29)
![\displaystyle\sum_{n=1}^k3n^2=(k+1)^3-1-3\frac{k(k+1)}2-k](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ek3n%5E2%3D%28k%2B1%29%5E3-1-3%5Cfrac%7Bk%28k%2B1%29%7D2-k)
![\displaystyle\sum_{n=1}^kn^2={k(k+1)(2k+1)}6](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5Ekn%5E2%3D%7Bk%28k%2B1%29%282k%2B1%29%7D6)
Now consider the cases where
is either odd or even.
- If
is odd, we can write
, where
. Then
![\displaystyle\sum_{m=1}^{25}(-1)^{2m-1}(2m-1)^2=-\sum_{m=1}^{25}(4m^2-4m+1)=-\frac{2\cdot25\cdot26\cdot51}3+2\cdot25\cdot26-25](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5E%7B25%7D%28-1%29%5E%7B2m-1%7D%282m-1%29%5E2%3D-%5Csum_%7Bm%3D1%7D%5E%7B25%7D%284m%5E2-4m%2B1%29%3D-%5Cfrac%7B2%5Ccdot25%5Ccdot26%5Ccdot51%7D3%2B2%5Ccdot25%5Ccdot26-25)
![\displaystyle\sum_{m=1}^{25}(-1)^{2m-1}(2m-1)^2=-20,825](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5E%7B25%7D%28-1%29%5E%7B2m-1%7D%282m-1%29%5E2%3D-20%2C825)
- If
is even, we can write
and so
![\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=\sum_{m=1}^{25}4m^2=\frac{2\cdot25\cdot26\cdot51}3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5E%7B25%7D%28-1%29%5E%7B2m%7D%282m%29%5E2%3D%5Csum_%7Bm%3D1%7D%5E%7B25%7D4m%5E2%3D%5Cfrac%7B2%5Ccdot25%5Ccdot26%5Ccdot51%7D3)
![\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=22,100](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bm%3D1%7D%5E%7B25%7D%28-1%29%5E%7B2m%7D%282m%29%5E2%3D22%2C100)
The original sum is obtained by adding the odd- and even-indexed sums together:
![\displaystyle\sum_{n=1}^{50}(-1)^nn^2=-20,825+22,100=\boxed{1275}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B50%7D%28-1%29%5Enn%5E2%3D-20%2C825%2B22%2C100%3D%5Cboxed%7B1275%7D)
Answer: put it in the microwave
Step-by-step explanation:
Answer:
sin18/cos72= sin(18) sec(72)
Step-by-step explanation:
rewrite it as a product: sin(18)/1 * 1/cos(72)
Divide sin(18) by 1 and rewrite 1/cos(72) as sec(72): sin (18) sec (72)