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Principle of Mathematical Induction

Long Answer Type

1.
Use principle of mathematical induction to prove that:

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

Let P(n): 1 + 2 + 3 + ......... + n = $space space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction$

I. For n = 1,

    P(1) : 1 = $fraction numerator 1 left parenthesis 1 plus 1 right parenthesis over denominator 2 end fraction rightwards double arrow space 1 space equals space 1 space rightwards double arrow space space straight P left parenthesis 1 right parenthesis$ is true.

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

      i.e. P(m): $1 plus 2 plus 3 plus........ space plus straight m space equals space fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction$           ....(i)

III.    For n = m + 1,

        P(m + 1): 1 + 2 + 3 + ........ + (m + 1) = $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$

or [1 + 2 + 3 + ...... + m] + (m + 1) = $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$

         [From (i), 1 + 2 + 3 + ...... + m = $fraction numerator straight m left parenthesis straight m plus 1 right parenthesis over denominator 2 end fraction$ ]

∴        P (m + 1): $space fraction numerator straight m left parenthesis straight m space plus space 1 right parenthesis over denominator 2 end fraction space plus space left parenthesis straight m space plus space 1 right parenthesis space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$

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

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

$rightwards double arrow$ $fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis over denominator 2 end fraction$

    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 $space space straight n element of straight N.$

761 Views

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

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

596 Views

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 $space space straight n element of straight N.$

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