Spooky + scary = Summation

Algebra Level 5

If $S_{k}, (k=1,2,3 \cdots ,100),$ denotes the infinite geometric progression sum , whose first term is $\dfrac{k - 1}{k!}$ and common ratio is $\dfrac{1}{k}$ , then find $\large \dfrac{(100)^{2}}{100!} + \displaystyle \sum_{k=2}^{100} | (k^{2} - 3 k + 1) \cdot S_{k} |$

The answer is 3.

1 solution

Rishabh Jain
Mar 17, 2016

Just loved solving this question @Mohit Gupta -Do you design them on your own or are they from a book??

Using formula for infinite GP:- $S_k=\dfrac{\frac{k-1}{k!}}{1-\frac 1k}=\dfrac{1}{(k-1)!}$

$\large\text{Let } \mathfrak{T}= \displaystyle \sum_{k=2}^{100} | (k^{2} - 3 k + 1) \cdot S_{k} |$ $\large =\displaystyle \sum_{k=2}^{100} |\dfrac{ (k^{2} - 3 k + 1) } {(k-1)!}|$

$\large =\displaystyle \sum_{k=2}^{100} |\dfrac{(k-1)(k-2)-1}{(k-1)!}|$

$\large=|-1|+\underbrace{\large \displaystyle \sum_{k=3}^{100} |\dfrac{1}{(k-3)!}-\dfrac{1}{(k-1)!}|}_{\mathfrak{W}}$

$\Large \mathfrak{W}=\left( \begin{aligned} & ~~~\left(1-\cancel{\color{#3D99F6}{\dfrac{1}{2!}}}\right)\\&+\left(1-\cancel{\color{#D61F06}{ \dfrac{1}{3!}}}\right)\\&+\left(\cancel{\color{#3D99F6}{\dfrac{1}{2!}}}-\cancel{\color{#EC7300}{\dfrac{1}{4!}}}\right) \\&\cdots\\&\cdots\\&\cdots\\&+\left(\cancel{\color{#20A900}{\dfrac{1}{96!}}}-\dfrac{1}{98!}\right)\\&+\left(\cancel{\color{#624F41}{\dfrac{1}{97!}}}-\dfrac{1}{99!}\right)\end{aligned} \right)$ $\color{#302B94}{\mathbf{A ~Telescopic ~Series}}$

$\large=2-\dfrac{1}{98!}-\dfrac{1}{99!}=2-\dfrac{100^2}{100!}$

Therefore required sum:- $\large\dfrac{100^2}{100!}+1+\mathfrak{W}$

$\Huge=\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{3}}}}}}$

No actually all these question are from different sources...the ones which i feel are cool so i modify them to harder level and then post on brilliant.....For example my this question Generalisations make things better was initially a easy question so i thought why not generalise it for all polynomials......other one Double refraction or trouble refraction ...was an easy question from INJSO....so i just post the problems which i think are worth sharing and more because of my curiosity to know that is a better soln. than mine possible for it :)

Mohit Gupta - 5 years, 3 months ago

Nice.... So how did you solved this one?

Rishabh Jain - 5 years, 3 months ago

Exactly like the one you solved.....however it took me just a bit time to discover that it wad a telescoping series.....

Please can you help me with this question $x^{2} + \frac{x^{2}}{(x+1)^{2} = 3$ .

Plz tell can we solve this question through algebraic manipulations...or some other comcept.

Mohit Gupta - 5 years, 3 months ago

@Mohit Gupta – I think you're saying : $x^2+\dfrac{x^2}{(x+1)^2}=3$ Make the substitution $x+1=t$ . $\implies (t-1)^2+\dfrac{(t-1)^2}{t^2}=3$ $\implies t^2-2t+\dfrac{1}{t^2}-\dfrac{2}{t}=1$ $\implies t^2+\dfrac{1}{t^2}-2(t+\dfrac 1t)=1$ $(t+\dfrac 1t)^2-2(t+\dfrac 1t)=3$ Again substitute $t+\dfrac 1t=p$ to get a quadratic in $p$ . $p^2-2p-3=0$ $(p-3)(p+2)=0$ Following process is easy now I think.... :-)

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – one thing.....in last eqn. you copied 1 as 3...

Mohit Gupta - 5 years, 3 months ago

@Mohit Gupta – No .. We have to add 2 on both the sides to transform $t^2+\dfrac{1}{t^2}$ to $(t+\dfrac 1t)^2$ .. Isn't it??

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – Ohhhhh.....i just forgot about that......thanks understood whete i was getting it wrong.....really a foolish mistake by me :P

Mohit Gupta - 5 years, 3 months ago

@Mohit Gupta – No problem.. :-) Keep giving beautiful problems...

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – Same problem as posted by mudit bansal, under the name 'Cool Summation'

A Former Brilliant Member - 5 years, 2 months ago

@A Former Brilliant Member – Ohk.. Thanks Might be this problem... But I solved mohit's problem first time and I loved it... Its a bit tricky..