The battle of 100

Algebra Level 2

∑ x = 1 100 [ ( x + 1 ) 100 − x 100 ] = ? \large{\sum^{100}_{x = 1} \left [ (x+1)^{100} - x^{100} \right] = \ ?}

10 1 100 − 1 101^{100} - 1 101 101 10 0 100 100^{100} 10 0 100 − 1 100^{100} - 1

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Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Chew-Seong Cheong
Oct 31, 2015

∑ x = 1 100 ( ( x + 1 ) 100 − x 100 ) = ∑ x = 1 100 ( x + 1 ) 100 − ∑ x = 1 100 x 100 = ∑ x = 2 101 x 100 − ∑ x = 1 100 x 100 = 10 1 100 − 1 100 = 10 1 100 − 1 \begin{aligned} \sum_{x=1}^{100} \left((x+1)^{100} - x^{100}\right) & = \sum_{\color{#3D99F6}{x=1}}^{\color{#3D99F6}{100}} (\color{#3D99F6}{x+1})^{100} - \sum_{x=1}^{100} x^{100} \\ & = \sum_{\color{#3D99F6}{x=2}}^{\color{#3D99F6}{101}} \color{#3D99F6}{x}^{100} - \sum_{x=1}^{100} x^{100} \\ & = 101^{100} - 1^{100} = \boxed{101^{100} - 1} \end{aligned}

12 Helpful 0 Interesting 0 Brilliant 0 Confused

i didn't understand

Adnan Malik - 5 years, 7 months ago

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Just expand the summations out then you should understand.

∑ x = 1 100 ( ( x + 1 ) 100 − x 100 ) = ∑ x = 1 100 ( x + 1 ) 100 − ∑ x = 1 100 x 100 = 2 100 + 3 100 + 4 100 + . . . + 10 1 100 − 1 100 + 2 100 + 3 100 + . . . + 10 0 100 = 10 1 100 − 1 100 \begin{aligned} \sum_{x=1}^{100} \left((x+1)^{100} - x^{100}\right) & = \color{#3D99F6}{\sum_{x=1}^{100} (x+1)^{100}} - \color{#D61F06}{\sum_{x=1}^{100} x^{100}} \\ & = \color{#3D99F6}{2^{100}+3^{100}+4^{100}+...+101^{100}} - \color{#D61F06}{1^{100} + 2^{100}+3^{100}+...+100^{100}} \\ & = \color{#3D99F6}{101^{100}} - \color{#D61F06}{1^{100}} \end{aligned}

Chew-Seong Cheong - 5 years, 7 months ago

2 100 + 3 100 + . . . + 10 0 100 + 10 1 100 − 1 100 − 2 100 − 3 100 − . . . − 10 0 100 2^{100}+3^{100}+...+100^{100}+101^{100}-1^{100}-2^{100}-3^{100}-...-100^{100} 10 1 100 − 1 \boxed{101^{100}-1}

Adrianus Felix - 5 years, 7 months ago

Can you please make it a bit more simpler. Thanks in advance.

Debmeet Banerjee - 5 years, 7 months ago

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∑ x = 1 100 ( ( x + 1 ) 100 − x 100 ) = ∑ x = 1 100 ( x + 1 ) 100 − ∑ x = 1 100 x 100 = 2 100 + 3 100 + 4 100 + . . . + 10 1 100 − 1 100 + 2 100 + 3 100 + . . . + 10 0 100 = 10 1 100 − 1 100 \begin{aligned} \sum_{x=1}^{100} \left((x+1)^{100} - x^{100}\right) & = \color{#3D99F6}{\sum_{x=1}^{100} (x+1)^{100}} - \color{#D61F06}{\sum_{x=1}^{100} x^{100}} \\ & = \color{#3D99F6}{2^{100}+3^{100}+4^{100}+...+101^{100}} - \color{#D61F06}{1^{100} + 2^{100}+3^{100}+...+100^{100}} \\ & = \color{#3D99F6}{101^{100}} - \color{#D61F06}{1^{100}} \end{aligned}

Chew-Seong Cheong - 5 years, 7 months ago

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Although I didn't understand, but thanks for making an effort.

Debmeet Banerjee - 5 years, 7 months ago

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