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Round one

1+1=2 2+2=4 4+4=8 8+8=16 16+16=32 32+32=64 64+64=128 128+128=256 256+256=512 2^0+2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8=? =2^1+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =2^2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 =2^3+2^3+2^4+2^5+2^6+2^7+2^8 =2^4+2^4+2^5+2^6+2^7+2^8 =2^5+2^5+2^6+2^7+2^8 =2^6+2^6+2^7+2^8 =2^7+2^7+2^8 =2^8+2^8 =2^9 = 512 ∵$ $∴$ $science$ $reason ∴ 2^0+2^0+2^1+2^2+2^3...... +2^{n} = 2^{n+1}

Round two

s
t1
t3
t4 t2
s=(1,1)$ , $t1=(2,4)$ , $t2=(4,2)$ , $t3=(3,3)$ , $t4=(4 ,1)

distand of s and t4 is 3.

s
1
2
t4

distand of s and t3 is 2 \times \sqrt2.

s
\sqrt2
t3
$∴ c^2=1+1=2 ∴ c=\sqrt2

distand of s and t2 is \sqrt10 \approx 3.162277.

s
t2

We can found though a^2+b^2=c^2 can get.

∵ a^2+b^2=c^2 ∴ (4-1)^2+(1-2)^2=c^2 ∴ 10=c^2 ∴ c=\sqrt10

Yeah,so distand of s and t2 is \sqrt10 \approx 3.162277 too!

∵$ $∴$ $science$ $reason ∴ distance=\sqrt{({{x_1}-{x_2}})^2+({{y_1}-{y_2}})^2}

Round 3

This is a circle.

problem: ∠BAC=？