An algebra problem by Fermin Cabrieto

Algebra Level 1


The answer is 5050.

Fermin Cabrieto by Fermin Cabrieto , 24, Philippines

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1 solution

We have:

1 + 4 + 9 + … + 10000 \quad\sqrt{1}+\sqrt{4}+\sqrt{9}+\ldots+\sqrt{10000}

= 1 + 2 + 3 + … + 100 =1+2+3+\ldots+100

= 100.101 2 = 5050 =\dfrac{100.101}{2}=\boxed{5050}

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that's right :)

Fermin Cabrieto - 5 years, 10 months ago

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Can somebody explain why for me please?

Anh Dinh - 5 years, 9 months ago

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BECAUSE if the squareroot 1 is equal to 1 then squareroot of 4 is 2 if you remember that if you continue adding 1 to 100 the same answer is the result and the sollution is :

=[100(100+1)]/2 =[100(101)]/2 =50(101) =5050

Fermin Cabrieto - 5 years, 9 months ago

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