find-C-0-100-2-C-2-100-2-C-4-100-2-C-6-100-2-C-100-100-2-

Question Number 157628 by mr W last updated on 25/Oct/21

f i n d

{(C_{0}^{100})}^{2} + {(C_{2}^{100})}^{2} + {(C_{4}^{100})}^{2} + {(C_{6}^{100})}^{2} + \dots + {(C_{100}^{100})}^{2} = ?

Answered by mindispower last updated on 25/Oct/21

\underset{k = 0}{\sum^{100}} {(C_{2 k}^{100})}^{2}

{(1 + i x)}^{100} {(1 - i x)}^{100} = \underset{k = 0}{\sum^{100}} C_{100}^{k} {(i x)}^{k} \underset{n = 0}{\sum^{100}} C_{n}^{100} {(- i x)}^{n}

= {(1 + x^{2})}^{100} = \underset{k = 0}{\sum^{100}} C_{100}^{k} {(i x)}^{k} . \underset{n = 0}{\sum^{100}} C_{n}^{100} {(- i x)}^{n}

n + k = 100

\Rightarrow C_{100}^{50} x^{100} = \underset{k = 0}{\sum^{100}} C_{100}^{k} {(i x)}^{k} . C_{100 - k}^{100} {(- i)}^{100 - k} x^{100 - k}

\Rightarrow \underset{k = 0}{\sum^{100}} C_{k}^{100} . C_{100 - k}^{100} {(- 1)}^{k} x^{100}

\Rightarrow \underset{k = 0}{\sum^{100}} {(- 1)}^{k} {(C_{k}^{100})}^{2} = C_{50}^{100}

{(1 + x)}^{100} {(1 + x)}^{100} = \underset{k = 0}{\sum^{100}} C_{k}^{100} x^{k} \underset{n = 0}{\sum^{100}} C_{n}^{100} x^{n}

C_{100}^{200} x^{100} = \underset{k = 0}{\sum^{100}} {(C_{k}^{100})}^{2}

Σ {(C_{2 k}^{100})}^{2} = \frac{1}{2} \underset{k = 0}{\sum^{100}} ({(C_{k}^{100})}^{2} + {(- 1)}^{k} {(C_{k}^{100})}^{2})

\underset{k = 0}{\sum^{50}} (C_{2 k}^{100}) = \frac{1}{2} (C_{100}^{200} + C_{50}^{100})

Commented by mr W last updated on 26/Oct/21

g r e a t! t h a n k s s i r!

Commented by mindispower last updated on 26/Oct/21

w i t h e p l e a s u r s i r

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