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3 years ago

PLZ CHECK, DON'T SPAM PLZ its attached below

Mathematics

2 answers:

sashaice [31]3 years ago

8 0

Answer:

Step-by-step explanation:

The constant term in a polynomial expression is the numeric term with no variable attached to it.

In this case the constant is + 1 → C

Oksanka [162]3 years ago

7 0

Answer:

correct answer given

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3 years ago

D²y/dx²=12x-12. express y in terms of x

Liono4ka [1.6K]

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3 years ago

3xy^2 what is the degree of the polynomial?

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3 years ago

Use mathematical induction to prove

Alex17521 [72]

$Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\$

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

$1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\$

Since the formula has been proven with n=1 and n=k+1, it is true. $\square$

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2 years ago