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Binomial Theorem

Short Answer Type

81. Find the term independent of x in the expansion of:

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82. Find the term independent of x in the expansion of:

open parentheses 2 straight x squared minus 1 over straight x close parentheses to the power of 12

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83. Find the term independent of x in the expansion of:

open parentheses 2 straight x minus straight x squared over 4 close parentheses to the power of 9

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Long Answer Type

84.
The absolute term in the expansion of: $<pre>uncaught exception: mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56 #0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>$ is 405. Find value of k.

Let $straight T subscript straight r plus 1 end subscript$ be the absolute term (term independent of x) in the expansion of $space open parentheses square root of straight x minus straight k over straight x squared close parentheses to the power of 10.$
$rightwards double arrow$ $straight T subscript straight r plus 1 end subscript$ contains $straight x to the power of 0.$
Now, $straight T subscript straight r plus 1 end subscript space equals space straight C presuperscript 10 subscript straight r space left parenthesis square root of straight x right parenthesis to the power of 10 minus straight r end exponent space open parentheses fraction numerator negative straight k over denominator straight x squared end fraction close parentheses to the power of straight r$

Power of $straight x equals open parentheses fraction numerator 10 minus straight r over denominator 2 end fraction close parentheses minus left parenthesis 2 straight r right parenthesis space equals space fraction numerator 10 minus 5 straight r over denominator 2 end fraction$

Power of x required = 0

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∴ $space space straight T subscript straight r plus 1 end subscript space equals space straight T subscript 3$ is the absolute term.

$space space straight T subscript 3 space equals space straight C presuperscript 10 subscript 2 space left parenthesis square root of straight x right parenthesis to the power of 8. space open parentheses fraction numerator negative straight k over denominator straight x squared end fraction close parentheses squared space equals space straight C presuperscript 10 subscript 2 space straight x to the power of 4. space straight k squared over straight x to the power of 4 equals space straight C presuperscript 10 subscript 2 space straight k squared$

∴ By condition of the question $straight C presuperscript 10 subscript 2 space straight k squared space equals space 405$
$rightwards double arrow$ $45 straight k squared space equals space 405$
$rightwards double arrow$ $space straight k squared space equals space 9$
Hence, $space space straight k equals plus-or-minus 3$

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85. Show that the co-efficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the co-efficients of two middle terms in the expansion of (1 + x)^2n-1

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86.

Find a, b and n in the expansion of $left parenthesis straight a plus straight b right parenthesis to the power of straight n$ if the first three terms of the expansion
are 729, 7290 and 30375, respectively.