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ONLY NEED TO SOLVE B, A IS DONE!

Suppose you add consecutive odd integers starting at 1.
1+3=4
1+3+5=9
1+3+5+7=16
a) Write a conjecture about your results.
When you add consecutive integers starting at one, your answer will be a perfect square.
b) Show that your conjecture is valid.

Respuesta :

chetankachana

Answer:

I assume that you already have the conjecture written down. If not, the conjecture is "The sum of consecutive odd integers starting at 1 is always a perfect square"

Proof: ↓↓↓↓

Step-by-step explanation:

We can start with the numbers given

1 + 3 = 4

[tex]\hookrightarrow[/tex]  4 is a perfect square = 2²

1 + 3 + 5 = 9

[tex]\hookrightarrow[/tex] 9 is a perfect square = 3²

1 + 3 + 5 + 7 = 16

[tex]\hookrightarrow[/tex]  16 is a perfect square = 4²

We can continue further:

1 + 3 + 5 + 7 + 9 = 25

[tex]\hookrightarrow[/tex] 25 is a perfect square = 5²

-Chetan K

mhanifa

Answer:

  • See below

Step-by-step explanation:

Consecutive odd integers make an AP with the first term a = 1 and common difference d = 2.

We see the given sums are perfect squares.

Lets show the sum is always a perfect square.

Sum of the first n terms is:

  • Sₙ = (2a + (n- 1)d)*n/2 =
  •       (2 + 2(n - 1)) * n/2 =
  •       2n*n/2 = n²

As we see the sum of the first n terms is the square of n.

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