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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.

The given series is:    $space space fraction numerator 1 over denominator 1.2 end fraction plus fraction numerator 1 over denominator 2.3 end fraction plus fraction numerator 1 over denominator 3.4 end fraction plus.......$

Let T_n be its nth term $straight T subscript straight n space equals space fraction numerator 1 over denominator left square bracket nth space term space of space 1 comma space 2 comma space 3 comma space........ right square bracket space left square bracket nth space term space of comma space 2 comma space 3 comma space 4 comma space.......... right square bracket end fraction$

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

$rightwards double arrow$    $space space space space space straight T subscript straight n space equals space 1 over straight n minus fraction numerator 1 over denominator straight n plus 1 end fraction$

Let $space space straight S subscript straight n$ be the sum of n terms of the series:

$straight S subscript straight n space equals space sum from straight k equals 1 to straight n of straight T subscript straight k space equals space sum from straight k equals 1 to straight n of open parentheses 1 over straight k minus fraction numerator 1 over denominator straight k plus 1 end fraction close parentheses space equals space open parentheses 1 over 1 minus 1 half close parentheses plus open parentheses 1 half minus 1 third close parentheses plus open parentheses 1 third minus 1 fourth close parentheses plus.... plus open parentheses 1 over straight n minus fraction numerator 1 over denominator straight n plus 1 end fraction close parentheses space equals space 1 minus fraction numerator 1 over denominator straight n plus 1 end fraction$

= $fraction numerator straight n plus 1 minus 1 over denominator straight n plus 1 end fraction$

$rightwards double arrow$ $space space space straight S subscript straight n space equals space fraction numerator straight n over denominator straight n plus 1 end fraction$

DEDUCTION: $straight S subscript straight n space equals space space fraction numerator begin display style straight n over straight n end style over denominator begin display style straight n over straight n end style plus begin display style 1 over straight n end style end fraction space equals space fraction numerator 1 over denominator 1 plus begin display style 1 over straight n end style end fraction comma space 1 over straight n$   approach to zero as n approach to infinity

∴ $straight S subscript infinity space equals space fraction numerator 1 over denominator 1 plus 0 end fraction space equals space 1$

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