Question

Determine the value of

Answer 1

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 $n$-th term of the sum, where $n=1,2,3,\ldots$, is

$(-1)^{n-1}(100+n)(100-n)=(-1)^n(n^2-100)$

We want to compute the sum,

$(101\cdot99)-(102\cdot98)+\cdots+(149\cdot51)-(150\cdot50)=\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)$

We have

$\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\sum_{n=1}^{50}(-1)^nn^2-100\sum_{n=1}^{50}(-1)^n$

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

$\displaystyle\sum_{n=1}^{50}(-1)^n(n^2-100)=\boxed{\sum_{n=1}^{50}(-1)^nn^2}$

In case you're not familiar with the formula for the sum of consecutive squares, we can derive it here. Recall that

$\displaystyle\sum_{n=1}^k1=k$

$\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2$

Notice that

$(n+1)^3-n^3=(n^3+3n^2+3n+1)-n^3=3n^2+3n+1$

and that

$\displaystyle\sum_{n=1}^k((n+1)^3-n^3)=(2^3-1^3)+(3^2-2^3)+\cdots+(k^3-(k-1)^3)+((k+1)^3-k^3)$

$\implies\displaystyle\sum_{n=1}^k((n+1)^3-n^3)=(k+1)^3-1$

Then

$(k+1)^3-1=\displaystyle\sum_{n=1}^k(3n^2+3n+1)$

$\displaystyle\sum_{n=1}^k3n^2=(k+1)^3-1-3\frac{k(k+1)}2-k$

$\displaystyle\sum_{n=1}^kn^2={k(k+1)(2k+1)}6$

Now consider the cases where $n$ is either odd or even.

• If $n$ is odd, we can write $n=2m-1$, where $m=1,2,3,\ldots,25$. Then

$\displaystyle\sum_{m=1}^{25}(-1)^{2m-1}(2m-1)^2=-\sum_{m=1}^{25}(4m^2-4m+1)=-\frac{2\cdot25\cdot26\cdot51}3+2\cdot25\cdot26-25$

$\displaystyle\sum_{m=1}^{25}(-1)^{2m-1}(2m-1)^2=-20,825$

• If $n$ is even, we can write $n=2m$ and so

$\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=\sum_{m=1}^{25}4m^2=\frac{2\cdot25\cdot26\cdot51}3$

$\displaystyle\sum_{m=1}^{25}(-1)^{2m}(2m)^2=22,100$

The original sum is obtained by adding the odd- and even-indexed sums together:

$\displaystyle\sum_{n=1}^{50}(-1)^nn^2=-20,825+22,100=\boxed{1275}$