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3 years ago

Find the product. (2f – 1)^2 =

Mathematics

2 answers:

Marina CMI [18]3 years ago

7 0

Answer: 4f^2 -4f+1

Step-by-step explanation:

2f-1)^2

This is equal to

(2f-1) * (2f-1)

FOIL

first: 2f*2f = 4f^2

outer: 2f*-1 = -2f

inner: -1*2f = -2f

Last = -1*-1 = 1

Add them together: = 4f^2 -2f-2f+1

Combine like terms 4f^2 -4f+1

S_A_V [24]3 years ago

7 0

Answer:

(2f-1)(2f-1)

2f^2-2f-2f+1

2f^2-4f+1

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Prove that 1³+2³+....n³=n²(n+1)²/4. principle of mathematics induction

alex41 [277]

<h3>The Simplified Question:-</h3>

$\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}$

$\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

<h3>Solution:-</h3>

Let

$\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

For n=1

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2$

$\\ \sf\longmapsto P(1)=(1)^2$

$\\ \bf\longmapsto P(1)=1=1^3$

Let k be any positive integer.

$\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2$

We have to prove that p(k+1) is true.

consider

$\sf 1^3+2^3\dots k^3+(k+1)^3$

$\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}$

$\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}$

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Thus P(k+1) is true whenever P(k) is true.

Hence by the Principal of mathematical induction statement P(n) is true for $\bf n\epsilon N.$

Note:-

We can solve without simplifying the Question .I did it for clear steps and understanding .

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Hey there! :)

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Find the product. (2f – 1)^2 =​

Find the product. (2f – 1)^2 =