Finite Series

Algebra Level 3

1 ( 40 ) + 3 ( 38 ) + 5 ( 36 ) + … + 37 ( 4 ) + 39 ( 2 ) = ? \large 1(40)+3(38)+5(36)+\ldots+37(4)+39(2)=\ ?


The answer is 5740.

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Ikkyu San by Ikkyu San , 23, Thailand

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1 solution

Ikkyu San
May 3, 2015

From sequence 1 , 3 , 5 , … , 39 → 2 ( 1 ) − 1 , 2 ( 2 ) − 1 , 2 ( 3 ) − 1 , … , 2 ( 20 ) − 1 1,3,5,\ldots,39\rightarrow 2(1)-1,2(2)-1,2(3)-1,\ldots,2(20)-1

Thus, a n = 2 n − 1 ⇒ ( 1 ) \color{#D61F06}{a_n=2n-1}\Rightarrow\color{#D61F06}{(1)}

From sequence 40 , 38 , 36 , … , 2 → 42 − 2 ( 1 ) , 42 − 2 ( 2 ) , 42 − 2 ( 3 ) , … , 42 − 2 ( 20 ) 40,38,36,\ldots,2\rightarrow 42-2(1),42-2(2),42-2(3),\ldots,42-2(20)

Thus, a n = 42 − 2 n ⇒ ( 2 ) \color{#3D99F6}{a_n=42-2n}\Rightarrow\color{#3D99F6}{(2)}

so the both sequences have 20 terms.

From ( 1 ) and ( 2 ) \color{#D61F06}{(1)}\text{ and }\color{#3D99F6}{(2)} that is

1 ( 40 ) + 3 ( 38 ) + 5 ( 36 ) + … + 37 ( 4 ) + 39 ( 2 ) 1(40)+3(38)+5(36)+\ldots+37(4)+39(2)

= ∑ n = 1 20 ( 2 n − 1 ) ( 42 − 2 n ) = ∑ n = 1 20 ( − 4 n 2 + 86 n − 42 ) = − 4 ∑ n = 1 20 n 2 + 86 ∑ n = 1 20 n − ∑ n = 1 20 42 = − 4 ( ( 20 ) ( 20 + 1 ) ( 2 ( 20 ) + 1 ) 6 ) + 86 ( 20 ( 20 + 1 ) 2 ) − 42 ( 20 ) = − 4 ( ( 20 ) ( 21 ) ( 41 ) 6 ) + 86 ( 20 ( 21 ) 2 ) − 840 = − 4 ( 2870 ) + 86 ( 210 ) − 840 = − 11480 + 18060 − 840 = 5740 \begin{aligned}=&\ \displaystyle\sum_{n=1}^{20}\color{#D61F06}{(2n-1)}\color{#3D99F6}{(42-2n)}\\=&\ \displaystyle\sum_{n=1}^{20}(\color{magenta}{-4n^2}\color{#624F41}{+86n}\color{#69047E}{-42})\\=&\ -4\color{magenta}{\displaystyle\sum_{n=1}^{20}n^2}+86\color{#624F41}{\displaystyle\sum_{n=1}^{20}n}-\color{#69047E}{\displaystyle\sum_{n=1}^{20}}42\\=&\ -4\color{magenta}{\left(\dfrac{(20)(20+1)(2(20)+1)}{6}\right)}+86\color{#624F41}{\left(\dfrac{20(20+1)}{2}\right)}-42\color{#69047E}{(20)}\\=&\ -4\color{magenta}{\left(\dfrac{(20)(21)(41)}{6}\right)}+86\color{#624F41}{\left(\dfrac{20(21)}{2}\right)}-840\\=&\ -4\color{magenta}{(2870)}+86\color{#624F41}{(210)}-840\\=&\ -11480+18060-840=\boxed{5740}\end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nice work as usual. I love the addition of colours in your work. Note: Try replacing b n b_n with a n a_n for equation number two, else you would have a n = 2 n − 1 = 42 − 2 n a_n = 2n-1=42-2n .

Despite a good conventional approach, can you think of a simpler solution? Hint: 1 + 40 = 3 + 38 = 5 + 36 = … = 39 + 2 1+40=3+38=5+36=\ldots = 39+2 .

Double the expression:

1 ( 40 ) + 2 ( 39 ) + 3 ( 38 ) + 4 ( 37 ) + … + 39 ( 2 ) + 40 ( 1 ) = ∑ n = 1 40 n ( 41 − n ) = 41 ∑ n = 1 40 n − ∑ n = 1 40 n 2 = 41 ⋅ 40 ⋅ 41 2 − 40 ( 41 ) ( 2 ⋅ 40 + 1 ) 6 = 11480 \begin{aligned} && 1(40) + 2(39) + 3(38) + 4(37) + \ldots + 39(2) + 40(1) \\ & = & \displaystyle \sum_{n=1}^{40} n(41-n) \\& = & \displaystyle 41 \sum_{n=1}^{40} n - \sum_{n=1}^{40} n^2 \\& = & 41 \cdot \frac{40\cdot 41}{2} - \frac{40(41)(2\cdot40+1)}{6} = 11480 \end{aligned}

Divide the expression back by two: 11480 ÷ 2 = 5740 11480 \div 2 = \boxed{5740} .

Pi Han Goh - 6 years, 1 month ago

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