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

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.
Advertisement

 Multiple Choice QuestionsShort 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

 Multiple Choice QuestionsLong 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

 Multiple Choice QuestionsShort 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

 Multiple Choice QuestionsLong Answer Type

124.

Find the sum to n terms of the series:

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

161 Views

Advertisement

 Multiple Choice QuestionsShort 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

Advertisement

127.

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


The given series is: n. 1 + (n - 1) . 2 + (n - 2) . 3 + ............. + 1. n 

Let Tk denote its kth term

∴           Tk = [kth term of n, n - 1, .....] cross times [kth term of 1, 2, 3, ........]

                 = [n + (k - 1) (-1)] [1 + (k - 1)1] = (n - k + 1) (k) = nk - k2 + k

              Tk = (n + 1)k - k2

Let space space straight S subscript straight n  denote the sum of the series


∴          straight S subscript straight n space equals space sum from straight k equals 1 to straight n of straight T subscript straight k space equals space sum from straight k equals 1 to straight n of left square bracket left parenthesis straight n plus 1 right parenthesis straight k space minus space straight k squared right square bracket space space equals space left parenthesis straight n plus 1 right parenthesis sum from straight k equals 1 to straight n of straight k space minus space sum from straight k equals 1 to straight n of straight k squared


                                        = fraction numerator left parenthesis straight n plus 1 right parenthesis straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction minus fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 1 right parenthesis over denominator 6 end fraction space equals space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 6 end fraction left square bracket 3 straight n plus 3 minus 2 straight n minus 1 right square bracket
   
                                         =   fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis straight n plus 2 right parenthesis over denominator 6 end fraction

94 Views

Advertisement

 Multiple Choice QuestionsLong 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

Advertisement

 Multiple Choice QuestionsShort 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

Advertisement