Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Mathematics : 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.

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