Verified answer

1) 3/(6+x)^3

= 3(6+x)^(-3)

= 3 * [6(1 + x/6)]^(-3)

= (3/216) * (1 + x/6)^(-3)

= (1/72) * [1 + Σ(n = 1 to ∞) C(-3, n) (x/6)^n], via binomial series

= (1/72) * [1 + Σ(n = 1 to ∞) ((-3)(-4)...(-3-n+1)/n!) (1/6)^n x^n]

= (1/72) * [1 + Σ(n = 1 to ∞) ((-1)^n (3)(4)...(n+2)/n!) (1/6)^n x^n]

= (1/72) * [1 + Σ(n = 1 to ∞) (1/2) * ((n+2)!/n!) (-1/6)^n x^n]

= (1/72) * Σ(n = 0 to ∞) ((n+2)(n+1)/2) (-1/6)^n x^n.

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2) Note that (d/dx) arcsin x = 1/√(1 - x^2) = (1 + (-x^2))^(-1/2).

Using the binomial series,

1/√(1 - x^2)

= (1 + (-x^2))^(-1/2)

= 1 + Σ(n = 1 to ∞) C(-1/2, n) (-x^2)^n, via binomial series

= 1 + Σ(n = 1 to ∞) [(-1/2)(-1/2 - 1)...(-1/2 - n + 1) / n!] (-1)^n x^(2n)

= 1 + Σ(n = 1 to ∞) (-1)^n [(1/2)(3/2)...((2n-1)/2) / n!] (-1)^n x^(2n)

= 1 + Σ(n = 1 to ∞) [(1)(3)...(2n-1)] x^(2n)/(2^n * n!).

So, integrating both sides from 0 to x yields

arcsin x = x + Σ(n = 1 to ∞) [(1)(3)...(2n-1)] x^(2n+1)/(2^n * (2n+1) * n!).

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3) 9/(1+x)^(1/4)

= 9(1 + x)^(-1/4)

= 9 [1 + Σ(n = 1 to ∞) C(-1/4, n) x^n], via binomial series

= 9 + Σ(n = 1 to ∞) 9((-1/4)(-1/4 - 1)...(-1/4 - n + 1)/n!) x^n

= 9 + Σ(n = 1 to ∞) 9(-1)^n ((1)(5)...(4n-3)/4^n) x^n/n!

= 9 + Σ(n = 1 to ∞) 9(-1/4)^n ((1)(5)...(4n-3)) x^n/n!.

Note that after the n = 0 term, we have an alternating series (with x = 0.1).

So, the error after n terms is bounded above by the following term.

==> We want 9(1/4)^(n+1) ((1)(5)...(4n-3)(4n+1) (0.1)^(n+1)/(n+1)! < 0.0005

==> n = 3 (or higher).

So, 9/(1.1)^(1/4) = 8.788 (to three decimal places).

I hope this helps!

a million) start up with (a million + x)^n = sum(ok=0 to n) C(n,ok) x^ok. Differentiate the two sides: n(a million + x)^(n-a million) = sum(ok=a million to n) ok C(n,ok) x^(ok-a million). (we can start up the index at a million, as a results of fact the ok=0 time era is consistent until now the differentiation step.) Multiply the two sides by utilising nx: n^2 x(a million + x)^(n-a million) = sum(ok=a million to n) nk C(n,ok) x^ok. permit x = 3: 3n^2 * 4^(n-a million) = sum(ok=a million to n) (nk * 3^ok) C(n,ok). --------------------------------------... 2) replace x with x^2 interior the binomial theorem start up with (a million + x^2)^n = sum(ok=0 to n) C(n,ok) x^(2k). Multiply the two sides by utilising x: x(a million + x^2)^n = sum(ok=0 to n) C(n,ok) x^(2k+a million). Differentiate the two sides: a million(a million + x^2)^n + x * n(a million + x^2)^(n-a million)(2x) = sum(ok=0 to n) (2k+a million) C(n,ok) x^(2k). permit x = a million: 2^n + n * 2^n = sum(ok=0 to n) (2k+a million) C(n,ok) ==> (n + a million) * 2^n = sum(ok=0 to n) (2k+a million) C(n,ok). stable success!