Show that 33-12-32-3-1-

3^3 12 = 3^2+3+1 Now formula used : #160; a^3 b^3 = (a b)(a^2+b^2+ab) Sol : 3^3 1^32 = (3 1)(3^2+1^2+(3)(1))2 = 3^3+3+1 or = 3^3 13 1 = ...

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Finding Square of a Number Using Identity

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Question

Show that $\frac{3^{3} - 1}{2} = 3^{2} + 3 + 1$ .

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$1 + 3 + 3^{2} + . . . . . . + 3^{n - 1} = \frac{3^{n} - 1}{2}$

1 + 3 + 3^{2} + . . . . . 3^{n - 1} = \frac{3^{n} - 1}{2}

{(1 + \frac{2}{3} + \frac{6}{3^{2}} + \frac{10}{3^{3}} + \dots upto \infty)}^{{log}_{(0.25)} ⎛ ⎝ \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots upto \infty ⎞ ⎠}

l

l^{2}

{lim}_{n \to \infty} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{n}})

${lim}_{x \to \infty} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . . + \frac{1}{3^{n}})$

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