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Question: Show that \[\dfrac{{1 \times {2^2} + 2 \times {3^2} + ... + n \times {{\left( {n + 1} \right)}^2}}}{...

Show that 1×22+2×32+...+n×(n+1)212×2+22×3+...+n2×(n+1)=3n+53n+1\dfrac{{1 \times {2^2} + 2 \times {3^2} + ... + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + ... + {n^2} \times \left( {n + 1} \right)}} = \dfrac{{3n + 5}}{{3n + 1}}

Explanation

Solution

In the given question, we have been given to prove a formula. It consists of a different combination of product of a number and its successor’s square and their sum in the denominator and product of a number and its predecessor’s square and their sum in the numerator. we are going to solve it by applying the formula of summation on the general term, writing their formula and then solving them.

Formula used:
We are going to use the following formulae:
∑n=n(n+1)2\sum n = \dfrac{{n\left( {n + 1} \right)}}{2}, ∑n2=n(n+1)(2n+1)6\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} and ∑n3=(n(n+1)2)2\sum {{n^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}

Complete step-by-step answer:
We have to show that 1×22+2×32+...+n×(n+1)212×2+22×3+...+n2×(n+1)=3n+53n+1\dfrac{{1 \times {2^2} + 2 \times {3^2} + ... + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + ... + {n^2} \times \left( {n + 1} \right)}} = \dfrac{{3n + 5}}{{3n + 1}}.
First, let us solve the numerator:
1×22+2×32+...+n×(n+1)21 \times {2^2} + 2 \times {3^2} + ... + n \times {\left( {n + 1} \right)^2}
The nth{n^{th}} term of this expression is given by:
n(n+1)2=n(n2+1+2n)=n3+2n2+nn{\left( {n + 1} \right)^2} = n\left( {{n^2} + 1 + 2n} \right) = {n^3} + 2{n^2} + n
So, we have to simplify:
∑n3+2n2+n=∑n3+∑2n2+∑n\sum {{n^3} + 2{n^2} + n = \sum {{n^3}} + \sum {2{n^2}} + \sum n }
We are going to use the following formulae:
∑n=n(n+1)2\sum n = \dfrac{{n\left( {n + 1} \right)}}{2}, ∑n2=n(n+1)(2n+1)6\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} and ∑n3=(n(n+1)2)2\sum {{n^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}
So, we have:
=n(n+1)2+2×n(n+1)(2n+1)6+(n(n+1)2)2=n(n+1)2+2n(n+1)(2n+1)6+n2(n+1)24= \dfrac{{n\left( {n + 1} \right)}}{2} + 2 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} = \dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{{2n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}
Simplifying and solving,
n(n+1)2(1+2(2n+1)3+n(n+1)2)\dfrac{{n\left( {n + 1} \right)}}{2}\left( {1 + \dfrac{{2\left( {2n + 1} \right)}}{3} + \dfrac{{n\left( {n + 1} \right)}}{2}} \right)
Taking the LCM and adding,
n(n+1)2(6+4(2n+1)+3n(n+1)6)\dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{6 + 4\left( {2n + 1} \right) + 3n\left( {n + 1} \right)}}{6}} \right)
Opening the brackets,
n(n+1)2(3n2+3n+8n+4+66)=n(n+1)12(3n2+11n+10)=n(n+1)12(3n+5)(n+2)\dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{3{n^2} + 3n + 8n + 4 + 6}}{6}} \right) = \dfrac{{n\left( {n + 1} \right)}}{{12}}\left( {3{n^2} + 11n + 10} \right) = \dfrac{{n\left( {n + 1} \right)}}{{12}}\left( {3n + 5} \right)\left( {n + 2} \right)
So, the numerator is equal to
n(n+1)(n+2)(3n+5)12\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{12}} …(i)
Now, let us solve the denominator:
12×2+22×3+...+n2×(n+1){1^2} \times 2 + {2^2} \times 3 + ... + {n^2} \times \left( {n + 1} \right)
The nth{n^{th}} of this expression is given by:
n2(n+1)=n3+n2{n^2}\left( {n + 1} \right) = {n^3} + {n^2}
So, we have to simplify:
∑n3+n2=∑n3+∑n2\sum {{n^3} + {n^2} = \sum {{n^3}} + \sum {{n^2}} }
So, we have,
n(n+1)(2n+1)6+n2(n+1)24\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}
Taking common terms out,
n(n+1)2[2n+13+n(n+1)2]\dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{2n + 1}}{3} + \dfrac{{n\left( {n + 1} \right)}}{2}} \right]
Taking the LCM and adding,
n(n+1)2[4n+2+3n(n+1)6]=n(n+1)2[3n2+7n+26]=n(n+1)(n+2)(3n+1)12\dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{4n + 2 + 3n\left( {n + 1} \right)}}{6}} \right] = \dfrac{{n\left( {n + 1} \right)}}{2}\left[ {\dfrac{{3{n^2} + 7n + 2}}{6}} \right] = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}
So, the denominator is equal to,
n(n+1)(n+2)(3n+1)12\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}} …(ii)
Now, dividing (i) by (ii),
1×22+2×32+...+n×(n+1)212×2+22×3+...+n2×(n+1)=n(n+1)(n+2)(3n+5)12n(n+1)(n+2)(3n+1)12\dfrac{{1 \times {2^2} + 2 \times {3^2} + ... + n \times {{\left( {n + 1} \right)}^2}}}{{{1^2} \times 2 + {2^2} \times 3 + ... + {n^2} \times \left( {n + 1} \right)}} = \dfrac{{\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right)}}{{12}}}}{{\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 1} \right)}}{{12}}}}
=(3n+5)(3n+1)= \dfrac{{\left( {3n + 5} \right)}}{{\left( {3n + 1} \right)}}
Hence, proved.

Note: In the given question, we had been given an expression. We had to prove it. The expression consisted of sum of natural numbers, sum of squares of natural numbers, and sum of cubes of natural numbers. We solved it by using the formula of summation, applying the formula of these terms and then solving them. So, it is important that we know how we can solve such questions, the procedure basically, and we must also remember the formulae of different terms.