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Sequences and Series

Short Answer Type

121. Find the sum of the squares of first n natural numbers. Or

Evaluate

space space sum straight n squared space space space or space space space space sum from straight k equals 1 to straight n of space straight k squared

129 Views

Long Answer Type

122. Find the sum of the cubes of first n natural numbers. Or

Evaluate

space space sum straight n cubed space space or space space space sum from straight k space equals 1 to straight n of space straight k cubed

Also, show that

sum straight n cubed space equals space left parenthesis sum straight n right parenthesis squared

110 Views

Short Answer Type

123. Find the sum of n terms of the series whose nth term is

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

128 Views

Long Answer Type

124.

Find the sum to n terms of the series:

1 . 2 . 4 + 2. 3 . 7 + 3. 4. 10 +..............

161 Views

Short Answer Type

125. Find the sum of the cubes of first n odd natural numbers. Or Sum the series to n terms l³ + 3³ + 5³ +..........

90 Views

126.

Find the sum to n terms the series $fraction numerator 1 over denominator 1.2 end fraction plus fraction numerator 1 over denominator 2.3 end fraction plus fraction numerator 1 over denominator 3.4 end fraction plus...........$ and deduce the sum to infinity.

102 Views

127.

Sum of the series n . 1 + (n - 1). 2 + (n - 2) . 3 + ........ + 1. n

94 Views

Long Answer Type

128. Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of these integers.

Let n + 1, n + 2, .....(n + m) be m consecutive integers

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

$space space straight S subscript 3 space equals space left parenthesis straight n plus 1 right parenthesis cubed plus left parenthesis straight n plus 2 right parenthesis cubed plus......... plus left parenthesis straight n plus straight m right parenthesis cubed$

= $left square bracket 1 cubed plus 2 cubed plus.......... plus straight n cubed plus left parenthesis straight n plus 1 right parenthesis cubed plus.......... plus left parenthesis straight n plus straight m right parenthesis cubed right square bracket space minus space left square bracket 1 cubed plus 2 cubed plus..... straight n cubed right square bracket$

= $sum from straight k equals 1 to straight n plus straight m of straight k cubed space minus space sum from straight k equals 1 to straight n of straight k cubed$ $space space space space space space space space open square brackets because space space sum from straight k equals 1 to straight n of straight k cubed space equals space fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction close square brackets$

= $fraction numerator left parenthesis straight n plus straight m right parenthesis squared left parenthesis straight n plus straight m plus 1 right parenthesis squared over denominator 4 end fraction minus fraction numerator straight n squared left parenthesis straight n plus 1 right parenthesis squared over denominator 4 end fraction$

= $open square brackets fraction numerator left parenthesis straight n plus straight m right parenthesis left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction minus fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets space cross times space open square brackets fraction numerator left parenthesis straight n plus straight m right parenthesis space left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets$

$straight S subscript 1 space equals space left parenthesis straight n space plus space 1 right parenthesis space plus space left parenthesis straight n space plus space 2 right parenthesis space plus space.......... space plus space left parenthesis straight n space plus space straight m right parenthesis$

= [1 + 2 + 3 +........ + n + (n + 1) + ..... + (n + m)] - [1 + 2 + ....... + n]

$space space space space equals space sum from straight k equals 1 to straight n plus straight m of straight k space minus space sum from straight k equals 1 to straight n of straight k space equals space fraction numerator left parenthesis straight n plus straight m right parenthesis space left parenthesis straight n plus straight m plus 1 right parenthesis over denominator 2 end fraction space minus space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction$ $open square brackets because space space sum from straight k equals 1 to straight n of straight k space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction close square brackets$

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

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

which is an integer because (n + m) (n + m + 1) and n(n + 1) are both even numbers being the product of two consecutive integers. Hence, the sum of cubes of any number of consecutive integers is divisible by the sum of these integers.

154 Views

Short Answer Type

129.

Show that $fraction numerator 1 cross times 2 squared plus 2 cross times 3 squared plus...... plus straight n cross times left parenthesis straight n plus 1 right parenthesis squared over denominator 1 squared cross times 2 plus 2 squared cross times 3 plus...... plus straight n squared cross times left parenthesis straight n plus 1 right parenthesis end fraction space equals space fraction numerator 3 straight n plus 5 over denominator 3 straight n plus 1 end fraction$

157 Views

130.

Find the sum to n terms of the following series:

5² + 6² + 7² + ....+ 20².

137 Views