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Principle of Mathematical Induction

Long Answer Type

Use principle of mathematical induction to prove that:

$1 space plus space 2 space plus space 3 space plus space... space plus space straight n space equals space fraction numerator straight n left parenthesis straight n space plus space 1 right parenthesis over denominator 2 end fraction$

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2.
Prove the following by using the principle of mathematical induction for all $straight n space element of space straight N colon$

$1 cubed plus 2 cubed plus 3 cubed plus space... space plus space straight n cubed space equals space open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets squared$

Let $straight P left parenthesis straight n right parenthesis space colon space 1 cubed plus 2 cubed plus 3 cubed plus..... plus straight n cubed space equals space open square brackets fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets squared$

I. For n = 1,
 $straight P left parenthesis 1 right parenthesis colon space 1 cubed space equals space open square brackets fraction numerator 1 left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction close square brackets squared space rightwards double arrow 1 space equals space 1 space rightwards double arrow space straight P left parenthesis 1 right parenthesis$ is true.

II. Suppose the statement is true for n = m, $straight m space element of space straight N$

 i.e., "<pre ... (i)

III. For n = m + 1,

 $straight P left parenthesis straight m plus 1 right parenthesis colon space 1 cubed plus 2 cubed plus 3 cubed plus......... plus left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

or $left square bracket 1 cubed space plus space 2 cubed space plus space 3 cubed space plus space....... space plus space straight m cubed right square bracket space plus space left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

 From (i), $1 cubed plus 2 cubed plus 3 cubed plus space............ space plus space straight m cubed space equals space open square brackets fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction close square brackets squared$

∴ $space space straight P left parenthesis straight m plus 1 right parenthesis space colon space open square brackets fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction close square brackets squared space plus space left parenthesis straight m plus 1 right parenthesis cubed space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

$rightwards double arrow space space left parenthesis straight m plus 1 right parenthesis squared space open square brackets straight m squared over 4 plus left parenthesis straight m plus 1 right parenthesis close square brackets space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

$rightwards double arrow space space space left parenthesis straight m plus 1 right parenthesis squared open parentheses fraction numerator straight m squared plus 4 straight m plus 4 over denominator 4 end fraction close parentheses space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

$rightwards double arrow space space space fraction numerator left parenthesis straight m plus 1 right parenthesis squared left parenthesis straight m plus 2 right parenthesis squared over denominator 4 end fraction space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared space rightwards double arrow space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared space equals space open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction close square brackets squared$

which is true

∴ P(m + 1) is true

∴ P(m) is true $rightwards double arrow$ P(m + 1) is true.

Hence, by mathematical induction, P(n) is true for all $straight n element of space straight N.$

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Prove the following by using the principle of mathematical induction for all $straight n element of straight N$ :

$1 plus 3 plus 3 squared plus....... space plus 3 to the power of straight n minus 1 end exponent space equals space fraction numerator 3 to the power of straight n minus 1 over denominator 2 end fraction$

398 Views

Prove the following by using the principle of mathematical induction for all $straight n element of straight N$ :

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

376 Views

Prove the following by using the principle of mathematical induction for all $straight n element of straight N$ .

$fraction numerator 1 over denominator 3.5 end fraction space plus space fraction numerator 1 over denominator 5.7 end fraction space plus space fraction numerator 1 over denominator 7.9 space end fraction space plus space......... space plus space fraction numerator 1 over denominator left parenthesis 2 straight n plus 1 right parenthesis left parenthesis 2 straight n plus 3 right parenthesis end fraction space equals space fraction numerator straight n over denominator 3 space left parenthesis 2 straight n space plus space 3 right parenthesis end fraction$

329 Views

Prove the following by principle of mathematical induction for all $straight n element of straight N$ :

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

226 Views

Prove the following by using the principle of mathematical induction for all $space space space space straight n element of straight N.$

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

221 Views

Prove the following by using the principle of mathematical induction for all $space space straight n element of straight N.$

$1.3 space plus space 2.3 squared space plus space 3.3 cubed space plus space........ space plus space straight n.3 to the power of straight n space equals space fraction numerator left parenthesis 2 straight n minus 1 right parenthesis space 3 to the power of straight n plus 1 end exponent space plus space 3 over denominator 4 end fraction$

189 Views

Prove the following by using the principle of mathematical induction for all $straight n element of straight N colon$

$space space 1 plus fraction numerator 1 over denominator 1 plus 2 end fraction plus fraction numerator 1 over denominator 1 plus 2 plus 3 end fraction plus....... plus fraction numerator 1 over denominator 1 plus 2 plus 3 plus........ plus straight n end fraction space equals space fraction numerator 2 straight n over denominator straight n plus 1 end fraction$

173 Views

10.

Prove the following by using the principle of mathematical induction for all $space space straight n element of straight N.$

$1 plus 3 plus 5 plus........... space plus space left parenthesis 2 straight n minus 1 right parenthesis space equals space straight n squared$

276 Views

Previous Year Papers

Test Series

Test Yourself

Long Answer Type

2. Prove the following by using the principle of mathematical induction for all