#17 of March 2015 Grade 11 - Natural Sciences & Liberal Arts CSAT (Korean SAT) Mock test

Algebra Level 4

A sequence { a n } \{a_n\} satisfies a 1 = 1 a_1=1 and a n + 1 n + 2 = a n n + 1 2 \dfrac{a_{n+1}}{n+2}=\dfrac{a_n}{n}+\dfrac{1}{2} for all natural numbers n . n.

Let S = 1 3 + 1 4 + 1 5 + ⋯ + 1 102 . S = \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\cdots+\dfrac{1}{102}.

Find the value of a 103 − 5356 S . \boxed{a_{103}-5356S}.

This problem has been modified by me, and is a part of <Grade 11 CSAT Mock test> series .


The answer is 8086.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Aug 2, 2017

Divide both sides by n + 1 n+1 and you see that

a n + 1 ( n + 1 ) ( n + 2 ) = a n n ( n + 1 ) + 1 2 ( n + 1 ) . \dfrac{a_{n+1}}{(n+1)(n+2)}=\dfrac{a_n}{n(n+1)}+\dfrac{1}{2(n+1)}.

Then define a n n ( n + 1 ) = b n . \dfrac{a_n}{n(n+1)}=b_n.

b n + 1 = b n + 1 2 ( n + 1 ) , b_{n+1}=b_n+\dfrac{1}{2(n+1)}, and b 1 = a 1 2 = 1 2 . b_1=\dfrac{a_1}{2}=\dfrac{1}{2}.

Substitute n = 1 , 2 , 3 , ⋯ , k − 1 n=1,~2,~3,~\cdots,~k-1 and then add them sides by sides.

b k = b 1 + 1 2 ( 1 2 + 1 3 + 1 4 + ⋯ + 1 k ) = 1 2 ( 1 + 1 2 + 1 3 + ⋯ + 1 k ) ( k ≥ 2 ) \begin{aligned} b_k &=b_1+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+~ \cdots ~+\dfrac{1}{k}\right) \\ &=\dfrac{1}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+~ \cdots ~+\dfrac{1}{k}\right)~(k\ge2) \end{aligned}

We see that this also holds for k = 1 , k=1, as b 1 = 1 2 . b_1=\dfrac{1}{2}.

Therefore,

a n = n ( n + 1 ) b n = n ( n + 1 ) 2 ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) . a_n=n(n+1)b_n=\dfrac{n(n+1)}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+~\cdots~+\dfrac{1}{n}\right).

a 103 = 103 × 104 2 ( 1 + 1 2 + 1 3 + ⋯ + 1 103 ) = 5356 ( 1 + 1 2 + S + 1 103 ) a_{103}=\dfrac{103\times104}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+~\cdots~+\dfrac{1}{103}\right)=5356\left(1+\dfrac{1}{2}+S+\dfrac{1}{103}\right)

∴ a 103 − 5356 S = 5356 ( 1 + 1 2 + 1 103 ) = 5356 + 2678 + 52 = 8086 . \therefore~a_{103}-5356S=5356\left(1+\dfrac{1}{2}+\dfrac{1}{103}\right)=5356+2678+52=\boxed{8086}.

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What is meant by adding "sides by sides"? Also I'm not understanding how u got that step just after adding sides by sides...can u please elaborate it little bit and give?

Skanda Prasad - 3 years, 8 months ago

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b 2 = b 1 + 1 2 ⋅ 2 b 3 = b 2 + 1 2 ⋅ 3 b 4 = b 3 + 1 2 ⋅ 4 b 5 = b 4 + 1 2 ⋅ 5 . . . b k = b k − 1 + 1 2 ⋅ k b_2=b_1+\dfrac{1}{2\cdot2} \\ b_3=b_2+\dfrac{1}{2\cdot3} \\ b_4=b_3+\dfrac{1}{2\cdot4} \\ b_5=b_4+\dfrac{1}{2\cdot5} \\ ... \\ b_k=b_{k-1}+\dfrac{1}{2\cdot k}

Add them all, and you get

b 2 + b 3 + b 4 + ⋯ + b k = b 1 + b 2 + b 3 + ⋯ b k − 1 + 1 2 ( 1 2 + 1 3 + 1 4 + ⋯ + 1 k ) b_2+b_3+b_4+~\cdots~+b_k=b_1+b_2+b_3+~\cdots b_{k-1}+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+~\cdots~+\dfrac{1}{k}\right)

Now we can eliminate b 2 + b 3 + b 4 + ⋯ + b k − 1 b_2+b_3+b_4+~\cdots~+b_{k-1} from each side.

b k = b 1 + 1 2 ( 1 2 + 1 3 + 1 4 + ⋯ + 1 k ) . \boxed{b_k=b_1+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+~\cdots~+\dfrac{1}{k}\right)}.

Boi (보이) - 3 years, 8 months ago

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Oh, ok... Thanks a lot! Great solution, though!

Skanda Prasad - 3 years, 8 months ago

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@Skanda Prasad – No problem, and thanks! ^^

Boi (보이) - 3 years, 8 months ago

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