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 [tex]n[/tex]-th term of the sum, where [tex]n=1,2,3,\ldots[/tex], is
[tex](-1)^{n-1}(100+n)(100-n)=(-1)^n(n^2-100)[/tex]
We want to compute the sum,
[tex](101\cdot99)-(102\cdot98)+\cdots+(149\cdot51)-(150\cdot50)=\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)[/tex]
We have
[tex]\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\sum_{n=1}^{50}(-1)^nn^2-100\sum_{n=1}^{50}(-1)^n[/tex]
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
[tex]\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\boxed{\sum_{n=1}^{50}(-1)^nn^2}[/tex]
In case you're not familiar with the formula for the sum of consecutive squares, we can derive it here. Recall that
[tex]\displaystyle\sum_{n=1}^k1=k[/tex]
[tex]\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2[/tex]
Notice that
[tex](n+1)^3-n^3=(n^3+3n^2+3n+1)-n^3=3n^2+3n+1[/tex]
and that
[tex]\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)[/tex]
[tex]\implies\displaystyle\sum_{n=1}^k((n+1)^3-n^3)=(k+1)^3-1[/tex]
Then
[tex](k+1)^3-1=\displaystyle\sum_{n=1}^k(3n^2+3n+1)[/tex]
[tex]\displaystyle\sum_{n=1}^k3n^2=(k+1)^3-1-3\frac{k(k+1)}2-k[/tex]
[tex]\displaystyle\sum_{n=1}^kn^2={k(k+1)(2k+1)}6[/tex]
Now consider the cases where [tex]n[/tex] is either odd or even.
[tex]\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[/tex]
[tex]\displaystyle\sum_{m=1}^{25}(-1)^{2m-1}(2m-1)^2=-20,825[/tex]
[tex]\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=\sum_{m=1}^{25}4m^2=\frac{2\cdot25\cdot26\cdot51}3[/tex]
[tex]\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=22,100[/tex]
The original sum is obtained by adding the odd- and even-indexed sums together:
[tex]\displaystyle\sum_{n=1}^{50}(-1)^nn^2=-20,825+22,100=\boxed{1275}[/tex]
