Question

please solve this question..​

Answer 1

Question :–

▪︎ Evaluate $\: \: \displaystyle \sum_{n=1}^{13} (i^n + i^{n+1}) \: \:$ , Where ${ \: \: \bold{ n \in \: N}} \: \:$

ANSWER :–

• Let's put –

$\\ \implies{ \bold{ \: \: p = }} \displaystyle \sum_{n=1}^{13} (i^n + i^{n+1}) \: \:$

• We show write this as –

$\\ \implies{ \bold{ \: \: p = }} \displaystyle \sum_{n=1}^{13} (i^n) + \displaystyle \sum_{n=1}^{13} (i^{n+1}) \: \:$

• Now put the values of 'n' –

$\\ \implies{ \bold{ \: \: p = \: }} ( i + {i}^{2} + ...... + {i}^{13}) + ( {i}^{2} + {i}^{3} + ...... + {i}^{14} ) \: \:$

• Properties of iota :–

$\\ \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: \: i = \sqrt{ - 1}$

$\\ \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: \: {i}^{2} = - 1$

$\\ \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: \: {i}^{3} = - i$

$\\ \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: \: {i}^{4} = 1$

$\\ \: \: \: \: \: \: \: \: { \huge{.}} \: \: \: \: i + {i}^{2} + {i}^{3} + {i}^{4} = 0$

• So that –

$\\ \implies{ \bold{ \: \: p = \: }} [ ( i + ...+ {i}^{4} )+ ({i}^{5} ... + {i}^{8}) + ( {i}^{9} + ... + {i}^{12} ) + {i}^{13} ] +[ ( {i}^{2} + ... + {i}^{5} ) + ( {i}^{6} + ... + {i}^{9} ) + ( {i}^{10} ... + {i}^{13} ) + {i}^{14} ] \: \:$

$\\ \implies{ \bold{ \: \: p = \: }} [ ( i + ...+ {i}^{4} )+ {i}^{4} ({i} + ... + {i}^{4}) + {i}^{8} ( {i}^{} + ... + {i}^{4} ) + {i}^{12} .i] +[ i( {i}^{} + ... + {i}^{4} ) + {i}^{5} ( {i}^{} + ... + {i}^{4} ) + {i}^{9} ( {i}^{} + ... + {i}^{4} ) + {i}^{12} .i] \: \:$

$\\ \implies{ \bold{ \: \: p = \: }} {i}^{12} .i + {i}^{12} . {i}^{2} \: \: \: \: \: [ \: \because \: \: i + {i}^{2} + {i}^{3} + {i}^{4} = 0 ]$

$\\ \implies{ \bold{ \: \: p = \: }}i + {i}^{2} \: \: \: \: \: [ \: \because \: \: {i}^{12} = {i}^{4} . {i}^{4} . {i}^{4} = 1 ]$

$\\ \implies{ \bold{ \: \: p = \: }} \: i - 1 \: \: \: \: \: [ \: \because \: \: {i}^{2} = - 1 ]$

Hence , $\\ \: \: \displaystyle \sum_{n=1}^{13} (i^n + i^{n+1}) = i - 1 \\ \: \:$

Answer 2

Answer:

mujhe zoom meeting ke bare mein kuch nahi pata.... sorry