Sequences and Series Test - 20

Since, $$S_{ n }$$ denote the sum of cubes of the first $$n$$ natural numbers.
Therefore, $$S_{ n }=\dfrac { { n }^{ 2 }{ \left( n+1 \right)  }^{ 2 } }{ 4 } $$
Since, $$s_{ n }$$ denote the sum of the first n natural numbers.
Therefore, $$s_{ n }=\dfrac { { n }{ \left( n+1 \right)  } }{ 2 } $$

Now, $${ T }_{ n }=\sum _{ r=1 }^{ n } \dfrac { S_{ r } }{ s_{ r } } =\sum _{ r=1 }^{ n } \dfrac { r\left( r+1 \right)  }{ 2 } $$

$$\Rightarrow { T }_{ n }=\dfrac { 1 }{ 2 } \left\{ \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\sum _{ r=1 }^{ n } r \right\} =\dfrac { { n }{ \left( n+1 \right)  }{ \left( 2n+1 \right)  } }{ 12 } +\dfrac { { n }{ \left( n+1 \right)  } }{ 4 } $$

$$\Rightarrow { T }_{ n }=\dfrac { n\left( n+1 \right) \left( n+2 \right)  }{ 6 } $$

$${ a }_{ n+1 }={ a }_{ n }+4n+3$$

$$a_1=1, a_{n+1}^{2}=(2\alpha^2-2\alpha+1)^n$$

Let $$S=\sum _{ r=1 }^{ n } \left\{ \left( 2r-1 \right) a+\dfrac { 1 }{ b^{ r } }  \right\} $$

$$\Rightarrow S=2a\sum _{ r=1 }^{ n } r-a\sum _{ r=1 }^{ n } +\sum _{ r=1 }^{ n } \left( \dfrac { 1 }{ b^{ r } }  \right) $$

$$\Rightarrow S=an\left( n+1 \right) -an+\dfrac { 1 }{ b } \left[ \dfrac { { \left( 1/b \right)  }^{ n }-1 }{ \left( 1/b \right) -1 }  \right] $$

$$\Rightarrow S=a{ n }^{ 2 }+\dfrac { { b }^{ n }-1 }{ { b }^{ n }\left( b-1 \right)  } $$

Ans: B

We find that
$$S=1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n \cdots (1)$$
$$\because (1+2+3+...+(n-1)+n)^{2}=1^{2}+2^{2}+3^{2}+...(n-1)^{2}+n^{2}$$
$$+2(1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n )$$    from $$(1)$$
$$(\sum n)^{2}=\sum n^{2}+2S$$
$$\Rightarrow S=\displaystyle \frac{(\sum n)^{2}-\sum n^{2}}{2}$$

$$=\displaystyle \frac{\left \{ \displaystyle \frac{n(n+1)}{2} \right \}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$$
$$=\displaystyle \frac { \displaystyle \frac{n^{2}(n+1)}{4}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$$
$$=\displaystyle \frac{n(n+1)}{24}\left \{ 3n(n+1)+2(2n+1) \right \}$$
$$=\displaystyle \frac{n(n+1)(3n^{2}-n-2)}{24}$$
$$=\displaystyle \frac{(n+1)n(n-1)(3n+2)}{24}$$

Label the table entry in the i th row and j th column by $$a_{ ij }$$
where the bottom-left corner is in the first row and first column.
Let  $$\displaystyle b_{ij}= 10(i-1)+j$$ be the number originally in the i th row and j th column. Observe that $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$ 
is invariant. Indeed, every time entries  
$$\displaystyle a_{mn}, a_{pq}, a_{rs}$$ are changed (with m+r= 2p and n+s= 2q), P increases or decreases by $$\displaystyle b_{mn}-2b_{pq}+b_{rs},$$ 
But this equals $$\displaystyle 10\left ( \left ( m-1 \right )+ \left ( r-1 \right )-2\left ( p-1 \right )+\left ( n+s-2q \right )\right )= 0.$$ 
In the beginning $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$ at the end, the entires $$a_{ij}$$ equal the $$b_{ij}$$ 
In some order,we now have $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
By the rearrangement inequality, this is at least  $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
with equality only when each  $$a_{ij}=b_{ij}$$
The equality does occur since P is invariant. Therefore the $$a_{ij}$$ do indeed equal the $$b_{ij}$$
in the same order, and thus the entries $$1, 2, ... , 100$$ appear in their original order.

The sequence can be written as :
$${ S }_{ n }={ n }^{ 3 }+{ \left( n-2 \right)  }^{ 3 }+{ \left( n-4 \right)  }^{ 3 }+......+{ 1 }^{ 3 }-\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ ={ n }^{ 3 }+{ \left( n-2 \right)  }^{ 3 }+{ \left( n-4 \right)  }^{ 3 }+......+{ 1 }^{ 3 }+\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2\left[ { \left( n-1 \right)  }^{ 3 }+{ \left( n-3 \right)  }^{ 3 }+{ \left( n-5 \right)  }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ =\sum _{ r=1 }^{ n }{ r^{ 3 } } -2\times { 2 }^{ 3 }\left[ { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+............+{ \left( \dfrac { n-1 }{ 2 }  \right)  }^{ 3 } \right] $$
we have taken 8 common from the 2nd part of the sequence.
$$=\sum _{ r=1 }^{ n }{ r^{ 3 } } -16\sum _{ 1 }^{ \dfrac { n-1 }{ 2 }  }{ r^{ 3 } } \\ =\left[ { \dfrac { n\left( n+1 \right)  }{ 2 }  }^{ 2 } \right] -16\left[ { \dfrac { \dfrac { n-1 }{ 2 } \left( \dfrac { n-1 }{ 2 } +1 \right)  }{ 2 }  }^{ 2 } \right] \\ ={ \dfrac { n\left( n+1 \right)  }{ 2 }  }^{ 2 }-16\left[ { \dfrac { \left( n-1 \right) \left( n+1 \right)  }{ 8 }  }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right)  }^{ 2 } }{ 4 } \left[ { n }^{ 2 }-{ \left( n-1 \right)  }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right)  }^{ 2 }\left( 2n-1 \right)  }{ 4 } $$

Let $$S=11+23+45+87+...$$up to $$n$$ terms
$$\Rightarrow \ S=\left( 5.2+1 \right) +\left( 5.{ 2 }^{ 2 }+3 \right) +\left( 5.{ 2 }^{ 3 }+5 \right) +\left( 5.{ 2 }^{ 4 }+7 \right) +...$$up to $$n$$ terms

Now, $${ a }_{ n }=5{ .2 }^{ n }+2n-1$$
Therefore, $$S=\sum _{ n=1 }^{ n }{ { a }_{ n } } =\sum _{ n=1 }^{ n }{ \left( 5{ .2 }^{ n }+2n-1 \right)  } $$
$$\Rightarrow S=5\sum _{ n=1 }^{ n }{ { 2 }^{ n } } +2\sum _{ n=1 }^{ n }{ n } -\sum _{ n=1 }^{ n }{ 1 } =\dfrac { 5.2\left( { 2 }^{ n }-1 \right)  }{ 2-1 } +\dfrac { 2n\left( n+1 \right)  }{ 2 } -n$$
$$\Rightarrow S=10\left( { 2 }^{ n }-1 \right) +{ n }^{ 2 }$$

Let, $$S=1^{ 3 }-2^{ 3 }+3^{ 3 }-4^{ 3 }+...+9^{ 3 }$$

$$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -\left( 2^{ 3 }+4^{ 3 }+6^{ 3 }+8^{ 3 } \right) $$

$$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -2^{ 3 }\left( 1^{ 3 }+2^{ 3 }+3^{ 3 }+4^{ 3 } \right) $$

$$\Rightarrow S=\sum _{ n=1 }^{ 5 }{ { \left( 2n-1 \right)  }^{ 3 } } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } } $$

$$\Rightarrow S=8\sum _{ n=1 }^{ 5 }{ { n }^{ 3 } } -12\sum _{ n=1 }^{ 5 }{ { n }^{ 2 } } +6\sum _{ n=1 }^{ 5 }{ { n } } -\sum _{ n=1 }^{ 5 }{ 1 } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } } $$

Consider
$$\displaystyle \left ( b_{1}+b_{2}+b_{3}+...+b_{n} \right )^{2}=b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}+2\sum_{i= j}b_ib_j $$
taking $$\displaystyle b_{1}=1,b_{2}=2,...,b_{n}=n$$
$$\displaystyle \therefore \left ( 1+2+3+...+n \right )^{2}=1^{2}+2^{2}+3^{2}+...+n^{2}+2\displaystyle \sum $$(Product of number taken two at a time)
$$\Rightarrow 2 \sum{b_ib_j} \displaystyle =\left ( 1+2+3+...+n \right )^{2}-\sum_{n=1}^{n}n^{2}\displaystyle=\frac{n^{2}\left ( n+1 \right )^{2}}{4}-\frac{n\left ( n+1 \right
)\left ( 2n+1 \right )}{6}=\frac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{12}$$
$$\Rightarrow \sum{b_ib_j} =\dfrac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{24} $$