Question

$\large \sf\red{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} ( \frac{5! \times 5}{4!} ) ^{ \frac{1}{2} } }$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: \sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} \bigg( \dfrac{5! \times 5}{4!} \bigg) ^{ \dfrac{1}{2} } }$

can be rewritten as

$\rm \: = \: \sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} \bigg( \dfrac{5 \times 4! \times 5}{4!} \bigg) ^{ \dfrac{1}{2} } }$

$\rm \: = \: \sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} \bigg( \dfrac{5 \times 5}{1} \bigg) ^{ \dfrac{1}{2} } }$

$\rm \: = \: \sf{ - \sum \limits_{i = 1}^{461}5 ( - 1)^{i}}$

$\rm \: = \: \sf{ - 5\sum \limits_{i = 1}^{461} ( - 1)^{i}}$

$\rm \: = \: - 5\bigg[ - 1 + {( - 1)}^{2} + {( - 1)}^{3} + - - - 461 \: terms \bigg]$

So, its a Geometric progression with first term, a = - 1 and common ratio, r = - 1 and number of terms, n = 461.

So, using sum of n terms of GP series, we get

$\rm \: = \: - 5\bigg[\dfrac{ - 1[ {( - 1)}^{461} - 1]}{ - 1 - 1} \bigg]$

$\rm \: = \: 5\bigg[\dfrac{ - 1 - 1}{ -2} \bigg]$

$\rm \: = \: 5\bigg[\dfrac{ - 2}{ -2} \bigg]$

$\rm \: = \: 5$

Hence,

$\rm :\longmapsto\: \boxed{\tt{ \: \sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} \bigg( \dfrac{5! \times 5}{4!} \bigg) ^{ \dfrac{1}{2} } } = 5 \: }}$

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Formula Used

↝ Sum of n  terms of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a( {r}^{n} - 1)}{r - 1} \: provided \: that \: r \: \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of GP..

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

Answer 2

Question:-

$\sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} ( \frac{5! \times 5}{4!} ) ^{ \frac{1}{2} } }$

Given:-

$\sf{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} ( \frac{5! \times 5}{4!} ) ^{ \frac{1}{2} } }$

Solution:-

$\sf \: S = \bigg [ { - \sum \limits_{i = 1}^{461} ( - 1)^{i} ( \frac{5! \times 5}{4!} ) ^{ \frac{1}{2} } } \bigg]$

$\sf \therefore \: S = - \sum \limits_{i = 1} ^{461} \bigg[( - 1) {}^{i } \: \: 5 \bigg] \sf = ( - 5) \bigg[ \sum \limits_{i = 1} ^{461} ( - 1) \: {}^{i} \bigg ]$

$\sf \therefore S = - ( - 5) \times [ - 1 + 1 - 1 + 1 - 1 + .... \: 461 \: terms ]$

$\sf \therefore \:S = ( - 5) \times [0 - 1] = ( - 5) \times ( - 1) = 5$

Answer:-

$\sf\red{ - \sum \limits_{i = 1}^{461} ( - 1)^{i} ( \frac{5! \times 5}{4!} ) ^{ \frac{1}{2} } = 5 }$

Hope you have satisfied. ⚘