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Principle of Mathematical Induction

Long Answer Type

Use principle of mathematical induction to prove that:

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

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 $straight n element of straight N$ :

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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

6.
Prove the following by principle of mathematical induction for all $straight n element of straight N$ :

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Let $straight P left parenthesis straight n right parenthesis colon space fraction numerator 1 over denominator 1.2.3 end fraction plus space fraction numerator 1 over denominator 2.3.4 end fraction plus fraction numerator 1 over denominator 3.4.5 end fraction plus space...... plus fraction numerator 1 over denominator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis straight n plus 2 right parenthesis end fraction space equals space fraction numerator straight n left parenthesis straight n plus 3 right parenthesis over denominator 4 left parenthesis straight n plus 1 right parenthesis left parenthesis straight n plus 2 right parenthesis end fraction$

I. For n = 1,

$straight P left parenthesis 1 right parenthesis colon space fraction numerator 1 over denominator 1.2.3 end fraction space equals space fraction numerator 1 left parenthesis 1 plus 3 right parenthesis over denominator 4 left parenthesis 1 plus 1 right parenthesis left parenthesis 1 plus 2 right parenthesis end fraction rightwards double arrow 1 over 6 space equals space fraction numerator 4 over denominator 4 left parenthesis 2 right parenthesis left parenthesis 3 right parenthesis end fraction rightwards double arrow space 1 over 6 space equals space 1 over 6$

∴ P(1) is true

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

∴     $straight P left parenthesis straight m right parenthesis colon space fraction numerator 1 over denominator 1.2.3 end fraction plus fraction numerator 1 over denominator 2.3.4 end fraction plus fraction numerator 1 over denominator 3.4.5 end fraction plus........ plus space fraction numerator 1 over denominator straight m left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction space equals space fraction numerator straight m left parenthesis straight m plus 3 right parenthesis over denominator 4 left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction space space space... left parenthesis i right parenthesis$

III. For n = m +1,

      $straight P left parenthesis straight m plus 1 right parenthesis colon space fraction numerator 1 over denominator 1.2.3 end fraction plus fraction numerator 1 over denominator 2.3.4 end fraction plus fraction numerator 1 over denominator 3.4.5 end fraction plus....... plus fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis left parenthesis straight m plus 3 right parenthesis end fraction equals fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 1 plus 3 right parenthesis over denominator 4 left parenthesis straight m plus 1 plus 1 right parenthesis left parenthesis straight m plus 1 plus 2 right parenthesis end fraction$

or
   $fraction numerator 1 over denominator 1.2.3 end fraction plus fraction numerator 1 over denominator 2.3.4 end fraction plus fraction numerator 1 over denominator 3.4.5 end fraction plus........ plus fraction numerator 1 over denominator straight m left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis left parenthesis straight m plus 3 right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space$

                                                                                              $space space equals fraction numerator left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 4 right parenthesis over denominator 4 left parenthesis straight m plus 2 right parenthesis left parenthesis straight m plus 3 right parenthesis end fraction$

From (i),

fraction numerator 1 over denominator 1.2.3 end fraction space plus space fraction numerator 1 over denominator 2.3.4 end fraction plus fraction numerator 1 over denominator 3.4.5 end fraction plus space........ plus space fraction numerator 1 over denominator straight m left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction equals space fraction numerator straight m left parenthesis straight m plus 3 right parenthesis over denominator 4 left parenthesis straight m plus 1 right parenthesis left parenthesis straight m plus 2 right parenthesis end fraction

∴ $straight P left parenthesis straight m plus 1 right parenthesis colon fraction numerator straight m left parenthesis straight m plus 3 right parenthesis over denominator 4 left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis end fraction equals fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis end fraction$

$rightwards double arrow space space fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis end fraction open square brackets fraction numerator straight m left parenthesis straight m plus 3 right parenthesis over denominator 4 end fraction plus fraction numerator 1 over denominator straight m plus 3 end fraction close square brackets space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis end fraction$

$rightwards double arrow$ $fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis end fraction open square brackets fraction numerator straight m cubed plus 6 straight m squared plus 9 straight m plus 4 over denominator 4 left parenthesis straight m plus 3 right parenthesis end fraction close square brackets equals fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis end fraction$

$rightwards double arrow space space space fraction numerator 1 over denominator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 2 right parenthesis end fraction open square brackets fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 space left parenthesis straight m plus 3 right parenthesis end fraction close square brackets space equals space fraction numerator left parenthesis straight m plus 1 right parenthesis space left parenthesis straight m plus 4 right parenthesis over denominator 4 space left parenthesis straight m plus 2 right parenthesis space left parenthesis straight m plus 3 right parenthesis end fraction$

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

∴ P(m + 1) is true

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

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

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

6. Prove the following by principle of mathematical induction for all :