If a- b- c are three consecutive positive integers- such that 1-2log ea -1-2log ec - 1-2ca-1 - 1-31-2ca-13 - 1-51-2ca-15 - - - log eb Find b

1/2log _ea +1/2log _ec + (12ca+1) + 13(12ca+1)^3 + 15(12ca+1)^5 + ... #160; #160; #160; #160; #160; = #160;12[ #160;log _ea + #16 ...

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Integration of Piecewise Continuous Functions

If a, b, c ...

Question

If $a, b, c$ are three consecutive positive integers, such that $\frac{1}{2} {log}_{e} a + \frac{1}{2} {log}_{e} c + (\frac{1}{2 c a + 1}) + \frac{1}{3} {(\frac{1}{2 c a + 1})}^{3} + \frac{1}{5} {(\frac{1}{2 c a + 1})}^{5} + . . . \infty = {log}_{e} b$
Find $b$

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If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

(iii) $\frac{(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}} = \frac{a + b + c}{a - b + c}$

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 4 a & 3 b & 2 c a & b & c \end{matrix} ∣ ∣ ∣ ∣ = 0

a, b, c

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

l o g_{e} (\frac{a + b}{2}) = \frac{1}{2} (l o g_{e} a + l o g_{e} b)

2 {log}_{e} n - {log}_{e} (n + 1) - {log}_{e} (n - 1) = \frac{1}{n^{2}} + \frac{1}{a n^{4}} + \frac{1}{3 n^{6}} + . . . \infty

a

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