Binomial (1 Viewer)

[azureus88]

Need help with 2 questions:

(i) If n is a positive integer, use the bionmial theorem to prove (5+sqrt11)^n +(5-sqrt11)^n is an integer.

(ii) Use bionomial expansions and the binomial theroem to find the value of (0.99)^13 to 5 siginificant figures.

btw, is there a better way to do the last question instead of literally expanding . thanks

 

[alakazimmy]

Need help with 2 questions:

(i) If n is a positive integer, use the bionmial theorem to prove (5+sqrt11)^n +(5-sqrt11)^n is an integer.

(ii) Use bionomial expansions and the binomial theroem to find the value of (0.99)^13 to 5 siginificant figures.

btw, is there a better way to do the last question instead of literally expanding . thanks

(5+sqrt(11))ⁿ = 5ⁿ + 5^n-1*11^1/2 + 5^n-2*11¹+ ... ... 11^n/2

(5-sqrt(11))ⁿ = 5ⁿ - 5^n-1*11^1/2 + 5^n-2*11¹+ ... ... +(-1)ⁿ11^n/2

So when you add the 2 together, the odd terms cancel out (the ones with square roots in them) and the even terms add up (the integers).
Hence, the sum is an integer.

(ii) No idea why need binomial for that. Just put the numbers into a calculator.

 

[Trebla]

For (ii), use the expansion (1 - 0.01)¹³

 

[pyrodude1031]

.99^13= (1-0.01)^13={13C0} (1)^13 (-0.01)^0 + {13C1} (1)^12 (-0.01)^1 + {13C2} (1)^11 (-0.01)^2 + {13C3} (1)^10 (-0.01)^3 + {13C4} (1)^9 (-0.01)^4 + ... + {13C13} (1)^0 (-0.01)^13
Since we only require to 5 significant figures, we ignore {13C4} onwards. Choice if after trial and error, from your 'feeling' regarding the small size of the (0.01)^n factor.

= {13C0} (-0.01)^0 + {13C1} (-0.01)^1 + {13C2} (-0.01)^2 + {13C3} (-0.01)^3 + {13C4} (-0.01)^4
= 1 + 13(-0.01) + 78(-0.01)^2 + 286(-0.01)^3 + 715(-0.01)^4
= 1 + -0.13 + 0.0078 + -0.000286 + 0.00000715
= 0.87752115
~= 0.87752 (5sf)

Sorry, I use 'feeling' with maths.

 

[azureus88]

Ok thanks for the input ppl.