Find the best answers to your questions with the help of IDNLearn.com's knowledgeable users. Whether your question is simple or complex, our community is here to provide detailed and trustworthy answers quickly and effectively.
Answer:
Step-by-step explanation:
To prove that the sum of the first n even +ve integers is:
[tex]\mathsf{2+4+6+8+ . . . +2n = n^2+ n }[/tex]
By using mathematical induction;
For n = 1, we get:
2n = 2 × 1 = 2
2 = 1² + 1 ----- (1)
∴ the outcome is true if n = 1
However, let assume that the result is also true for n = k
Now, [tex]\mathsf{2+4+6+8+. . .+2k = k^2 + k --- (2)}[/tex]
[tex]\mathsf{2+4+6+8+. . .+2k+2(k+1)}[/tex]
we can now say:
[tex]\mathsf{= (k^2 + k) + 2(k + 1)} \\ \\ \mathsf{= k^2 + k + 2k + 1}[/tex]
[tex]\mathsf{= (k^2 + 2k + 1) + (k + 1)}[/tex]
[tex]\mathsf{= (k + 1)^2 + (k + 1)}[/tex]
∴
[tex]\mathsf{2 + 4 + 6 + 8 + . . . + 2k + 2(k + 1) = (k + 1)^2 + (k + 1)}[/tex]
Thus, the result is true for n = m+1, hence we can posit that the result is also true for each value of n.
As such [tex]\mathsf{2+4+6+8+. . .+2n = n^2 + n }[/tex]