Please help?

Question

Binomial expansion:

Use the expansion of (x+y)^4 to evaluate (1.03)^4 .

I'm confused with this too:

The first term of an arithmetic progression as a and the common difference is d. If the 5th,9th and 16th term form a three term geometric progression with common ration r, find the value of d in terms of a.

Answer 1

Here is the answer to your second question. The fifth term in the A. P. is a + 4d, the ninth is a + 8d, and the sixteenth is
a + 15d. Since these are in G. P. we know that
a + 8d = (a + 4d)*r and a + 15d = (a + 8d)*r . From the first equation we get r = (a + 8d)/(a + 4d) and from the second, we get r = (a + 15d)/(a + 8d) . Equating the two expressions for r, and multiplying through by the denominators, we get
(a + 8d)^2 = (a + 15d)*(a + 4d) , or a^2 + 16ad + 64d^2 =
a^2 + 19ad + 60d^2. This simplifies to 16ad + 64d^2 =
19ad + 60 d^2 . If d = 0 , the A. P. is a sequence of constant terms, and all terms are in G. P. with r = 1 . If d is not zero, we cancel d from the orevious equation and find
16a + 64d = 19a + 60d, or a = (4/3)*d. That is your answer.

As a check we notice that the 5th, 9th, and 16th terms are
(16/3)*d , (28/3)*d , and (49/3)*d which are in G. P. with ratio r = 7/4 .

Answer 2

(x+y)^4 = x^4 + 4x³y + 6x²y² + 4xy³ + y^4

(1.03)^4
= (1+0.03)^4
= 1^4 + 4 ·1³ · 0.03+ 6 · 1² · 0.03² + 4 · 1 ·0.03³ + 0.03^4
= 1 + 4 · 0.03+ 6 * 0.03² + 4 · 0.03³ + 0.03^4
= 1 + 0.12 +0.0054 + 0.000 108 + 0.000 000 81
= 1.125 508 81

Bye and good luck !!

Answer 3

If you know the expansion of (X+Y)^4 then you can find the solution by simply substituting X=1 and Y=0.03 in the expansion!!!


It's that simple!!

Answer 4

Newton had told me:
(x+y)^4 =x^4+4x^3y+6x^2y^2+4xy^3+y^4

let x=1,y=0.03.
y^2 ,y^3,y^4 is insignificant .

we will get the approximate answer x^4+4x^3y

that is 1.12

Answer 5

(1.03)^4 = (1 + 0.03)^4
= 1 + 4*(0.03) + 6*(0.03)^2 + 4*(0.03)^3 + (0.03)^4

neglecting higher order terms you'll get

(1.03)^4 = 1.12 (approx.)

Answer 6

I tend to agree.

Answer 7

(1.03)^4
=(1+0.03)^4

=1.1
+4.1.(0.03)
+6.1.(0.03)^2
+4.1(0.03)^3
+1.1(0.03)^4