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Question Number 24704 by math solver last updated on 25/Nov/17

Commented by math solver last updated on 25/Nov/17

$sum of series upto^{'} n^{'} terms is ?$
Commented by prakash jain last updated on 25/Nov/17
$T_i =(.5)^3 ×(i+1)^3 S_n =Σ_(i=1) ^n (.5)^3 ×(i+1)^3 =(.5)^3 [Σ_(i=0) ^(n+1) i^3 −1] =(.5)^3 [{(((n+1)(n+2))/2)}^2 −1]$
$T_{i} = {(. 5)}^{3} \times {(i + 1)}^{3}$ $S_{n} = \underset{i = 1}{\sum^{n}} {(. 5)}^{3} \times {(i + 1)}^{3}$ $= {(. 5)}^{3} [\underset{i = 0}{\sum^{n + 1}} i^{3} - 1]$ $= {(. 5)}^{3} [{\frac{(n + 1) (n + 2)}{2}}^{2} - 1]$
Commented by math solver last updated on 25/Nov/17

$gracias .$
Commented by math solver last updated on 25/Nov/17

$and how you write the 3 rd step ?$
Commented by jota+ last updated on 25/Nov/17
${i+1}={2 3 4 5 ...} {j} ={1 2 3 4...} Σ_(j=1) ^(n+1) a_j −a(1)=Σ_(i=1) ^n a_i (Propiedad telescopica)$
${i + 1} = {2 3 4 5 . . .}$ ${j} = {1 2 3 4 . . .}$ $\underset{j = 1}{\sum^{n + 1}} a_{j} - a (1) = \underset{i = 1}{\sum^{n}} a_{i}$ $(P r o p i e d a d t e l e s c o p i c a)$
	Answered by jota+ last updated on 25/Nov/17

	$T_{i} = B_{i}^{3} = {(0.5 + 0.5 i)}^{3} =$ $= 0 . 5^{3} {(1 + i)}^{3}$
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