Get Answers to all your Questions

header-bg qa

26. Show that \frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}
 

Answers (1)

best_answer

To prove : 

                  \frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)} = \frac{3n+5}{3n+1}

the nth term of numerator =n(n+1)^2=n^3+2n^2+n

nth term of the denominator =n^2(n+1)=n^3+n^2

 RHS:\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)}..........................1

=\frac{\sum _{k=1}^{n} a_k}{\sum _{k=1}^{n} a_k}=\frac{\sum _{k=1}^{n} k^{3}+2k^2+k}{\sum _{k=1}^{n} (k^3+k^2)}

                  Numerator : 

                S_n=\sum _{k=1}^{n} k^3+2\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} k

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{2.n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1)

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+8n+4+6}{6} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+11n+10}{6} \right )

                =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+11n+10 \right )

               =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+5n+10 \right )

              =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+5(n+2) \right )

               =\left [ \frac{n(n+1)}{12} \right ] \left ((n+2)(3n+5) \right )               

             =\frac{n(n+1)(n+2)(3n+5)}{12}........................................................2

Denominator :

    S_n=\sum _{k=1}^{n} k^3+\sum _{k=1}^{n} k^2

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{6}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{6}

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{2n+1}{3})

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+4n+2}{6} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+7n+2}{6} \right )

                =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+7n+2 \right )

                 =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n^2+6n+n+2 \right )

                 =\left [ \frac{n(n+1)}{12} \right ] \left ( 3n(n+2)+1(n+2)\right )

                =\left [ \frac{n(n+1)}{12} \right ] \left ( (n+2)(3n+1)\right )

              =\left [ \frac{n(n+1)(n+2)(3n+1)}{12} \right ].....................................3

 

From equation 1,2,3,we have

\frac{1 \times 2 ^ 2 + 2 \times 3^2 + ... + n \times (n+1)^2}{1^2 \times 2 + 2 ^2 \times 3 + ...+ n^2 \times ( n+1)}    =\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}

                                                                                  =\frac{3n+5}{3n+1}

Hence, the above expression is proved.

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads