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Question Number 61111 by Tawa1 last updated on 29/May/19

Please what does the 2 on the C mean .

C_{1}^{2} + 2 C_{2}^{2} + 3 C_{3}^{2} + \dots + n C_{n}^{2} = \frac{(2 n - 1)!}{{[(n - 1)!]}^{2}}

Does the 2 on C mean square ? ?

I mean : {(C_{1})}^{2} + 2 {(C_{2})}^{2} + 3 {(C_{3})}^{2} + \dots + n {(C_{n})}^{2}

which is also

{(\overset{n}{} C_{1})}^{2} + 2 {(\overset{n}{} C_{2})}^{2} + 3 {(\overset{n}{} C_{3})}^{2} + \dots + n {(\overset{n}{} C_{n})}^{2}

I just want to know what the 2 on C represent . Thanks .

C_{1}^{2} + 2 C_{2}^{2} + 3 C_{3}^{2} + \dots + n C_{n}^{2} = \frac{(2 n - 1)!}{{[(n - 1)!]}^{2}}

Commented by perlman last updated on 29/May/19

C_{n}^{k} = \frac{n!}{k! (n - k)!}

w i t h e n! = n (n - 1) \dots . 2 \times 1

m o r g e n e r a l y x! = Γ (x + 1) g a m m a f u n c t i o n d e f i n d f o r a l l z \in C w i t h e / R e (z) > 0

Γ (x + 1) = x Γ (x) a n d Γ (1) = 1

Commented by Tawa1 last updated on 29/May/19

But sir, how come we have 3 . C_{3}^{2} which is 3 . \overset{2}{} C_{3} is this correct

Answered by tanmay last updated on 29/May/19

{(1 + x)}^{n} = c_{0} + c_{1} x + c_{2} x^{2} + \dots + c_{n} x^{n}

\frac{d}{d x} [{(1 + x)}^{n}] = c_{1} + 2 c_{2} x + 3 c_{3} x^{2} + \dots + {n c}_{n} x^{n - 1}

x \times n {(1 + x)}^{n - 1} = c_{1} x + 2 c_{2} x^{2} + 3 c_{3} x^{3} + . . + {n c}_{n} x^{n}

n o w {(1 + \frac{1}{x})}^{n} = c_{0} + \frac{c_{1}}{x} + \frac{c_{2}}{x^{2}} + \dots + \frac{c_{n}}{x^{n}}

n o w

(c_{1} x + 2 c_{2} x^{2} + . . + {n c}_{n} x^{n - 1}) \times (c_{0} + \frac{c_{1}}{x} + \frac{c_{2}}{x^{2}} + . . + \frac{c_{n}}{x^{n}})

t a k e t h e t e r m s i n d e p e n d e n t o f x

c_{1}^{2} + 2 c_{2}^{2} + 3 c_{3}^{2} + \dots + {n c}_{n}^{2}

l e f t h a n d s i d e

x \times n {(1 + x)}^{n - 1} \times {(1 + \frac{1}{x})}^{n}

= x \times n {(1 + x^{})}^{n - 1} \times \frac{{(1 + x)}^{n}}{x^{n}}

= n \times x^{1 - n} \times {(1 + x)}^{2 n - 1}

l e t (r + 1) t h t e r m o f {(1 + x)}^{2 n - 1} c o n t a i n s

x^{n - 1} s o w h e n m u l t i p l i e d b y x^{1 - n} r e s u l t i s

i n d e p e n d e n t o f x

2 n - 1_{C_{r} \times {(x)}^{r}}

x^{r} = x^{n - 1} s o r = n - 1

n o w

n \times 2 n - 1_{C_{n - 1}} \times x^{n - 1} \times x^{1 - n}

n \times \frac{(2 n - 1)!}{(n - 1)! \times n!}

= \frac{(2 n - 1)!}{(n - 1)! (n - 1)!}

s o

c_{1}^{2} + 2 c_{2}^{2} + 3 c_{3}^{2} + \dots + {n c}_{n}^{2} = \frac{(2 n - 1)!}{(n - 1)! \times (n - 1)!}

h e n c e p r o v e d

Commented by Tawa1 last updated on 29/May/19

Wow, God bless you sir . But that C^{2}, is the 2 square or another way to write \overset{2}{} C

Commented by Tawa1 last updated on 29/May/19

Ohh, i get sir . It is square . C_{1} \times C_{1} = C_{1}^{2}

Commented by tanmay last updated on 29/May/19

y e s c_{1} \times c_{1} = {(c_{1})}^{2} s q u a r e i h a v e a v o i d e d (.)

j u s t w r i t t e n a s c_{1}^{2}

Commented by Tawa1 last updated on 29/May/19

Ohh, great . Thanks sir . I appreciate

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