3

Playing with numbers, I construct following expression.

Can it be shown that $$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$

attempt

We can construct following, using finite calculus as

$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$

$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$

Let $4k-1=2^n-1$ so we can write above claim as

$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$

Which is equivalent to show

$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$

Here I'm stuck. Thanks

Alma Arjuna
  • 3,759
Pruthviraj
  • 2,707

4 Answers4

3

For $n\leftarrow 2$, the equality holds.

Let's assume it holds for $n\leftarrow k\in\Bbb{N}$, $k\ge 2$. Then

$$\color{black}{\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 = 2^{2k-1}}$$

We want to show that the equality holds for $k+1$, i.e., that

$$\color{red}{\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k+1}}$$

In fact

$\color{black}{\begin{aligned} \sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 &&=\\ \sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=2^{k-1}+1}^{2^k}(-1)^i(2i-1)^2 &=\\ \sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^{2^{k-1}+i}(2(2^{k-1}+i)-1)^2 &=\\ \sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^k+2i-1)^2 &=\\ \sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)+(2i-1)^2) &=\\ 2\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)) &=\\ 2(2^{2k-1}) &+ \underbrace{\sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k})}_0 + \sum_{i=1}^{2^{k-1}}(-1)^i(2^{k+1}(2i-1)) &=\\ 2^{2k} &+ 2^{k+1}\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1) & \end{aligned}}$

It's well known (and easy to show) that $\begin{aligned}\sum_{i=1}^{m}(-1)^i(2i-1) = (-1)^m m\end{aligned}$, so it is proven that

$$\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k} + 2^{k+1}2^{k-1} = 2^{2k+1}$$

Alma Arjuna
  • 3,759
1

Using the Hockey-stick identity:

$$S(n)=\sum_{k=0}^{n-1} (2k+1)^2=\sum_{k=0}^{n-1} \left(8{k+1\choose 2}+1\right)=8{n+1\choose 3}+n=\frac{n(2n+1)(2n-1)}{3}$$

$$T(n)=\sum_{k=0}^{n-1} (4k+1)^2=\sum_{k=0}^{n-1} \left( 16k^2+8k+1\right)=\sum_{k=0}^{n-1} \left(32{k+1\choose2}-8{k+1\choose1}+9\right)\\=32{n+1\choose3}-8{n+1\choose2}+9n=\frac n3(16n^2-12n-1)$$

Let $N=2^{n-2}$, then

$$-1^2+3^2-5^2+\dots+(2^n-1)^2\\=-1^2+3^2-5^2+\dots+(4N-1)^2\\=S(2N)-2T(N)\\=8N^2\\=2^{2n-1}$$

To answer the comment:

$$1^2+3^2+5^2+\dots+(4N-1)^2=\sum_{k=0}^{2N-1} (2k+1)^2=S(2N)$$

$$1^2+5^2+\dots+(4N-3)^3=1^2+5^2+\dots+(4(N-1)+1)^3=\sum_{k=0}^{N-1}(4k+1)^2=T(N)$$

Now, the first row minus twice the second row yields:

$$-1^2+3^2-5^2+\dots+(4N-1)^2=S(2N)-2T(N)$$

Jean-Claude Arbaut
  • 23,277
0

Hint: What is $\sum_{k=1}^N k^2 i^k$, where $i^2=-1$? Then take its imaginary part.

Solution:

The sum in question is $-\sum_{k=1}^N k^2 i^k=\frac12 ((i-1)i^N N^2 -2i^N-i^{N+1}+i)$.

When $N$ is a multiple of $4$ we have $i^4=1$ and so the imaginary part is $\frac12 N^2$.

When $N=2^{n}$ (so that the last term is $2^n-1$), the sum is thus $\frac12 2^{2n} = 2^{2n-1}$.

lhf
  • 216,483
0

One acxtually gets $$\sum_{k=1}^{n} (-1)^k (2k-1)^2=\frac{1}{2}[1-(-1)^n+ 4(-1)^n n^2]$$

Z Ahmed
  • 43,235