Explanation
$$1=1$$, here $$n=1\Rightarrow \dfrac {1(1+1)}{2}=1$$ $$1+2=3$$, here $$n=2\Rightarrow \dfrac {2(2+1)}{2}=3$$
$$1+2+3=6$$, here $$n=3\Rightarrow \dfrac {3(3+1)}{2}=6$$
$$1+2+3+4=10$$, here $$n=4\Rightarrow \dfrac {4(4+1)}{2}=10$$
$${ n }^{ th }$$ term is sum of $$n$$ terms
Therefore, $${ T }_{ n }=\dfrac { n(n+1) }{ 2 } $$
The top layer has $$(13\times 13)$$ balls the layer below it will have $$(14\times 14)$$ balls
We have $$18$$ layers
So the total number of balls
$$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$$
Sum of squares of first $$n$$ natural numbers is $$\dfrac { n(n+1)(2n+1) }{ 6 } \\ $$
$$\therefore N=$$ sum of first $$30$$ $$-$$ sum of first $$12$$
$$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\ $$
$$\Rightarrow 8000<N<9000$$
$$\sum _{ r=1 }^{ n }{ \left( { r }^{ 2 }+1 \right) r! } $$
$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { r }^{ 2 }+1+2r-2r ) r! }$$
$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ 2 }.r!-2r.r! ) }$$
$$\Rightarrow \sum_{ r=1 }^{ n }{ ( { ( r+1) }^{ }.(r+1)!-2r.r! ) }$$
$$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2-1 )( r+1 )!-2( ( r+1-1 )r! ) ] }$$
$$\Rightarrow \sum_{ r=1 }^{ n }{ [ ( r+2 )( r+1 )!-( r+1 )!-2( ( r+1 )r!-r! ) ] }$$
$$\Rightarrow \sum_{ r=1 }^{ n }{ \left [ ( r+2 )!-( r+1 )!-2.( r+1 )!+2.r!) \right] }$$
$$T_1=3!-2!- 2.2!+2.1!$$
$$T_2=4!-3!-2.3!+2.2!$$
$$T_3=5!-4!-2.4!+2.3!$$
.
$$T_n=(n+2)!-(n+1)!-2.(n+1)!+2.n!$$
$$-------------------$$
$$Sum=S_n=(n+2)!-2!-2(n+1)!+2.1!$$
$$S_n=(n+2)!-2(n+1)!-2+2$$
$$S_n=(n+1)![n+2-2]$$
$$S_n=n.(n+1)!$$
Correct answer is C
$$\cfrac { 5 }{ 7 } =\cfrac { { a }_{ 2 } }{ 2! } +\cfrac { { a }_{ 3 } }{ 3! } +\cfrac { { a }_{ 4 } }{ 4! } +\cfrac { { a }_{ 5 } }{ 5! } +\cfrac { { a }_{ 6 } }{ 6! } +\cfrac { { a }_{ 7 } }{ 7! } \\ \Rightarrow \cfrac { 5 }{ 7 } =\cfrac { 1 }{ 2 } \left( { a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \right) \\ \cfrac { 10 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 3 }{ 7 } ={ a }_{ 2 }+\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\left( \cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) $$
So, as expression is less than $$1$$
$${ a }_{ 2 }=1\\ \cfrac { 3 }{ 7 } =\cfrac { 1 }{ 3 } \left( { a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+.....\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ 1+\cfrac { 2 }{ 7 } ={ a }_{ 3 }+\cfrac { 1 }{ 4 } \left( { a }_{ 4 }+......\cfrac { 1 }{ 5 } \left( { a }_{ 5 }+\cfrac { 1 }{ 6 } \left( { a }_{ 6 }+\cfrac { { a }_{ 7 } }{ 7 } \right) \right) \right) \\ { a }_{ 3 }=1$$
Similarly
$${ a }_{ 4 }=1\quad \quad { a }_{ 7 }=2\\ { a }_{ 5 }=0\\ { a }_{ 6 }=4$$
So, Sum $${ a }_{ 7 }-1+1+1+4+2\\ =9$$
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