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Otrada [13]
3 years ago
6

Prove the statement holds for all positive integers: 2 + 4 + 6 + ... + 2n = n² + n

Mathematics
1 answer:
crimeas [40]3 years ago
6 0

Proving a relation for all natural numbers involves proving it for n = 1 and showing that it holds for n + 1 if it is assumed that it is true for any n.

The relation 2+4+6+...+2n = n^2+n has to be proved.

If n = 1, the right hand side is equal to 2*1 = 2 and the left hand side is equal to 1^1 + 1 = 1 + 1 = 2

Assume that the relation holds for any value of n.

2 + 4 + 6 + ... + 2n + 2(n+1) = n^2 + n + 2(n + 1)

= n^2 + n + 2n + 2

= n^2 + 2n + 1 + n + 1

= (n + 1)^2 + (n + 1)

This shows that the given relation is true for n = 1 and if it is assumed to be true for n it is also true for n + 1.

<span>By mathematical induction the relation is true for any value of n.</span>

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The required equation is 0.50\ p+0.75\ e= 3.50\\

Given, Alejandro bought $3.50 worth of pencils and erasers at the school store.

The cost of each pencil is $0. 50.

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Since He denotes no. of pencil by p and no. of eraser by e, so the required equation will be,

0.50\ p+0.75\ e= 3.50\\.

Hence the required equation is 0.50\ p+0.75\ e= 3.50\\

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Find the value of X by solving the blue equation

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3x + 2 = x + 8 solve for x
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3x+2=x+8
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2x-6=0
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Sheri invests $10,000 in a simple interest account. What interest rate must the account pay so she earns $2,000 in 5 years? * O
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