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Secondary 1 | Maths
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Help Clara Chua! Anyone can contribute an answer, even non-tutors.
I am a P5 but my math teacher wanted us to try this. I need to a good explanation so i can understand.
Thank you!
Date Posted: 5 years ago
Views: 363
The idea is to make use of the formula (a + b)² = a² + 2ab + b². You will learn this in secondary school.
n = 2010² + 2011²
Notice that 2011 = 2010 + 1.
So n = 2010² + (2010 + 1)²
We let a = 2010, and b = 1.
So (2010 + 1)² = 2010² + 2(2010)(1) + 1²
Then n = 2010² + 2010² + 2(2010)(1) + 1²
n = 2(2010)² + 2(2010) + 1
Multiply both sides by 2,
2n = 4(2010)² + 4(2010) + 2
So 2n - 1 = 4(2010)² + 4(2010) + 2 - 1
2n - 1 = 4(2010)² + 4(2010) + 1
2n - 1 = (2 x 2 x 2010 x 2010) + (2 x
2 x 2010) + 1
2n - 1 = (2010 x 2)(2010 x 2) + (2 x 4020) + 1
2n - 1 = (4020)² + 2(4020)(1) + 1
Notice here that we can apply the same formula again. In this case we let a = 4020 and b = 1.
2n - 1 = (4020 + 1)²
2n - 1 = (4021)²
√(2n - 1) = 4021
n = 2010² + 2011²
Notice that 2011 = 2010 + 1.
So n = 2010² + (2010 + 1)²
We let a = 2010, and b = 1.
So (2010 + 1)² = 2010² + 2(2010)(1) + 1²
Then n = 2010² + 2010² + 2(2010)(1) + 1²
n = 2(2010)² + 2(2010) + 1
Multiply both sides by 2,
2n = 4(2010)² + 4(2010) + 2
So 2n - 1 = 4(2010)² + 4(2010) + 2 - 1
2n - 1 = 4(2010)² + 4(2010) + 1
2n - 1 = (2 x 2 x 2010 x 2010) + (2 x
2 x 2010) + 1
2n - 1 = (2010 x 2)(2010 x 2) + (2 x 4020) + 1
2n - 1 = (4020)² + 2(4020)(1) + 1
Notice here that we can apply the same formula again. In this case we let a = 4020 and b = 1.
2n - 1 = (4020 + 1)²
2n - 1 = (4021)²
√(2n - 1) = 4021
Is the ans 4021
Yes. It is indicated in the last step of the working above
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