An algebra problem by Pankaj Joshi

Algebra Level 5

Let S k S_k ( k = 1 , 2 , 3 , . . . , 100 ) (k = 1,2,3,...,100) denote the sum of infinite Geometric Progression whose first term is k − 1 k ! \dfrac {k-1}{k!} and the common ratio is 1 k \dfrac{1}{k} . Then find the value of 10 0 2 100 ! \dfrac{100^2}{100!} + ∑ k = 1 100 ∣ ( k 2 − 3 k + 1 ) . S k ∣ \sum_{k=1}^{100} \lvert(k^2 - 3k +1) . S_k \rvert


The answer is 3.

Pankaj Joshi by Pankaj Joshi , 23, India

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1 solution

Pankaj Joshi
May 22, 2014

S k = k − 1 k ! ( 1 − 1 k ) S_k = \dfrac { k-1 }{ k!(1-\dfrac { 1 }{ k } ) }

An important thing to be noted here is that S k S_k is not defined for k = 1 k=1 .So the range for summation reduces to {from 2 − 100 2 - 100 }.

S k = 1 ( k − 1 ) ! S_k = \dfrac{1}{(k-1)!}

∑ k = 2 100 ∣ ( k 2 − 3 k + 1 ) ( k − 1 ) ! ∣ \sum _{ k=2 }^{ 100 }{ |\dfrac { (k^ 2-3k+1) }{ (k-1)! } } |

∑ k = 2 100 ∣ ( k − 1 ) 2 − k ( k − 1 ) ! ∣ = ∑ k = 2 100 ∣ ( k − 1 ) ( k − 2 ) ! − k ( k − 1 ) ! ∣ \sum _{ k=2 }^{ 100 }{ |\dfrac { (k-1)^ 2-k }{ (k-1)! } } | = \sum _{ k=2 }^{ 100 }{ |\dfrac { (k-1) }{ (k-2)! } } -\dfrac { k }{ (k-1)! } |

Notice that the quantity inside modulus is negative for k=2 and the rest of the are positive.

Putting k=2 separately, we get

1 + ∑ k = 3 100 ∣ ( k − 1 ) ( k − 2 ) ! − k ( k − 1 ) ! ∣ 1 +\sum _{ k=3 }^{ 100 }{ |\dfrac { (k-1) }{ (k-2)! } } -\dfrac { k }{ (k-1)! } | . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ) ..........................(1)

1 + V k − 1 − V k 1 + V_{k-1} - V_{k}

Putting k= 3 to 100 we get ,

1 + V 2 − V 3 + V 3 − V 4 + V 4 . . . V 99 − V 100 1 + V_2 - V_3 + V_3 - V_4 +V_4 ... V_{99} - V_{100}

The middle terms get cancelled and we are let with ,

1 − V 100 + V 2 1 -V_{100} + V_2

Putting V {100} and V 2 in ( 1 ) (1) we get,

∑ k = 2 100 ∣ ( k 2 − 3 k + 1 ) ( k − 1 ) ! ∣ \sum _{ k=2 }^{ 100 }{ |\dfrac { (k^ 2-3k+1) }{ (k-1)! } } | = 3 − 100 99 ! 3- \dfrac{100}{99!}

So 10 0 2 100 ! − 100 99 ! + 3 \dfrac{100^2}{100!} -\dfrac{100}{99!} + 3

The final answer is 3 \boxed{3}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Note that " S k S_k is not defined for k = 1 k=1 " is not a true statement, since your 'equation definition' of S k S_k only applies if ∣ 1 k ∣ < 1 | \frac{1}{k} | < 1 , by the GP formula.

Instead, you should say that S 1 = 0 S_1 = 0 , since the first term is 0, and hence all terms are 0.

Calvin Lin Staff - 7 years ago

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Yes, you are right but I don't get you, Where am I wrong?

Also, can you help me with the correct latex for modulus? I have used vertical lines in the above question.

Pankaj Joshi - 7 years ago

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The minor mistake is saying that " S k S_k is not defined for k = 1 k = 1 ". Instead, you should bear in mind that the GP formula doesn't apply, and hence calculate S 1 = 0 S_1=0 separately.

Calvin Lin Staff - 7 years ago

ah... PJ there you are good question though...

pranav jangir - 6 years, 10 months ago

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