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Maths-

General

Easy

Question

If $not stretchy sum from r equals 1 to n of r to the power of 4 end exponent equals l open parentheses n close parentheses comma blank$ then $blank not stretchy sum from r equals 1 to n of open parentheses 2 r minus 1 close parentheses to the power of 4 end exponent$ is equal to

$I (2 n) - I (n)$
$I (2 n) - 16 I (n)$
$I (2 n) - 8 I (n)$
$I (2 n) - 4 I (n)$

The correct answer is: $I open parentheses 2 n close parentheses minus 16 I left parenthesis n right parenthesis$

$I open parentheses 2 n close parentheses equals 1 to the power of 4 end exponent plus 2 to the power of 4 end exponent plus 3 to the power of 4 end exponent plus horizontal ellipsis plus open parentheses 2 n minus 1 close parentheses to the power of 4 end exponent plus open parentheses 2 n close parentheses to the power of 4 end exponent$
$equals left square bracket left parenthesis 1 to the power of 4 end exponent plus 3 to the power of 4 end exponent plus 5 to the power of 4 end exponent plus horizontal ellipsis plus open parentheses 2 n minus 1 close parentheses to the power of 4 end exponent right square bracket plus 2 to the power of 4 end exponent open parentheses 1 to the power of 4 end exponent plus 2 to the power of 4 end exponent plus 3 to the power of 4 end exponent plus 4 to the power of 4 end exponent plus horizontal ellipsis n to the power of 4 end exponent close parentheses$
$equals not stretchy sum from r equals 1 to n of open parentheses 2 r minus 1 close parentheses to the power of 4 end exponent plus 16 cross times I open parentheses n close parentheses$
$rightwards double arrow not stretchy sum from r equals 1 to n of open parentheses 2 r minus 1 close parentheses to the power of 4 end exponent equals I open parentheses 2 n close parentheses minus 16 I left parenthesis n right parenthesis$

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