Question

find the value of i^{42}+i^{47}+i^{56}+i^{29}​

Answer 1

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

Given expression is

$\rm :\longmapsto\: {i}^{42} + {i}^{47} + {i}^{56} + {i}^{29}$

Consider,

$\rm :\longmapsto\: {i}^{42}$

$\rm \:  =  \: {i}^{2 \times 21}$

$\rm \:  =  \: {( {i}^{2}) }^{21}$

$\rm \:  =  \: {( - 1)}^{21}$

$\rm \:  =  \: - 1$

$\rm\implies \:\boxed{\tt{ {i}^{42} = - 1}}$

Consider,

$\rm :\longmapsto\: {i}^{47}$

$\rm :\longmapsto\: {i}^{46} \times i$

$\rm \:  =  \: {i}^{2 \times 23} \times i$

$\rm \:  =  \: {( {i}^{2} )}^{23} \times i$

$\rm \:  =  \: {( - 1)}^{23} \times i$

$\rm \:  =  \: - i$

$\rm\implies \:\boxed{\tt{ {i}^{47} = - i}}$

Now, Consider

$\rm :\longmapsto\: {i}^{56}$

$\rm \:  =  \: {i}^{2 \times 28}$

$\rm \:  =  \: {( {i}^{2} )}^{28}$

$\rm \:  =  \: {( - 1)}^{28}$

$\rm \:  =  \: 1$

$\rm\implies \:\boxed{\tt{ {i}^{56} = 1}}$

Now, Consider

$\rm :\longmapsto\: {i}^{29}$

$\rm :\longmapsto\: {i}^{28} \times i$

$\rm \:  =  \: {( {i}^{2} )}^{28} \times i$

$\rm \:  =  \: {( - 1)}^{28} \times i$

$\rm \:  =  \: 1 \times i$

$\rm \:  =  \: i$

$\rm\implies \:\boxed{\tt{ {i}^{29} = i}}$

Now, Consider

$\rm :\longmapsto\: {i}^{42} + {i}^{47} + {i}^{56} + {i}^{29}$

$\rm \:  =  \: - 1 - i + 1 + i$

$\rm \:  =  \: 0$

Hence,

$\rm :\longmapsto\: {i}^{42} + {i}^{47} + {i}^{56} + {i}^{29} = 0$

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ADDITIONAL INFORMATION

$\rm :\longmapsto\:i = \sqrt{ - 1}$

$\rm :\longmapsto\: {i}^{2} = - 1$

$\rm :\longmapsto\: {i}^{3} = - i$

$\rm :\longmapsto\: {i}^{4} = 1$