Question

The sum of series is√3+3√2+6√3+. To 16th term is

Answer 1

Answer:

$1007769\sqrt{2} - 1007769\sqrt{3}$

Step-by-step explanation:

By looking at the series we can tell that it is a GP

Where $a = \sqrt{3}$, $r = \sqrt{6}$

We know that sum of GP = $S_n = \frac{a(r^{n}-1) }{r-1}$

Sum of 16 terms of the GP = $S_{16} = \frac{\sqrt{3} (\sqrt{6} ^{16} -1)}{\sqrt6 - 1}$

⇒ $S_{16} = \sqrt{3} \frac{(\sqrt{6}^{8})^{2} - (1^{8})^{2} }{\sqrt{6}-1 }$

⇒ $S_{16} = \sqrt{3} \frac{[(\sqrt{6} ^{8} -1^{8})( \sqrt{6} ^{8} +1^{8})]}{\sqrt{6} -1}$                                        ($a^{2} -b^{2} = (a+b)(a-b)$)

We'll keep applying the same formula till we get $\sqrt{6} -1$ in the numerator

⇒ $S_{16} = \sqrt{3} \frac{(\sqrt{6}-1)(\sqrt{6}+1)(\sqrt{6}^{2} +1^{2})(\sqrt{6}^{4} +1^{4})(\sqrt{6}^{8} +1^{8} )}{(\sqrt{6}-1)}$

⇒ $S_{16} = \sqrt{3}(\sqrt{6}+1)(\sqrt{6}^{2} +1^{2})(\sqrt{6}^{4} +1^{4})(\sqrt{6}^{8} +1^{8} )$

⇒ $S_{16}= \sqrt{3} (\sqrt{6}+1) (6+1) (36+1) (1296+1)$

⇒ $S_{16} = \sqrt{3} (\sqrt{6} -1)(7)(37)(1297)$

⇒ $S_{16} = \sqrt{3} (\sqrt{6} -1)(335923)$

⇒ $S_{16} = (3\sqrt{2} - \sqrt{3}) (335923)$

⇒ $S_{16} = 1007769\sqrt{2} - 1007769\sqrt{3}$