x^n এর সহগ নির্ণয় বিষয়ক

The value of C12+C22....+Cn2C_1 ^2+C_2 ^2....+C_n ^2 (where CiC_i is the ithi^{th} coefficient of (1+x)n(1+x)^n expansion), is:

হানি নাটস

(1+x)n=nC0+x(nC1)+x2(nC2)+x3(nC3)+...+xn(nCn){ \left( 1+x \right) }^{ n }=^{ n }{ { C }_{ 0 } }+x\left( ^{ n }{ { C }_{ 1 } } \right) { +{ x }^{ 2 }\left( ^{ n }{ { C }_{ 2 } } \right) + }{ x }^{ 3 }\left( ^{ n }{ { C }_{ 3 } } \right) +...+{ x }^{ n }\left( ^{ n }{ { C }_{ n } } \right) \\

(x+1)n=xn(nC0)+xn1(nC1)+...+x0(nCn){ \left( x+1 \right) }^{ n }={ x }^{ n }\left( ^{ n }{ { C }_{ 0 } } \right) +{ x }^{ n-1 }\left( ^{ n }{ { C }_{ 1 } } \right) +...+{ x }^{ 0 }\left( ^{ n }{ { C }_{ n } } \right)
Now, r=0n(nCr)2\quad \sum _{ r=0 }^{ n }{ { \left( ^{ n }{ { C }_{ r } } \right) }^{ 2 } } = co-efficient of xnx^n in (1+x)2n=2nCn(1+x)^{2n}=^{ 2n }{ { C }_{ n } }
Thus, 2nCn=(nC0)2+(nC1)2+...(nCn)2^{ 2n }{ { C }_{ n } }={ \left( ^{ n }{ { C }_{ 0 } } \right) }^{ 2 }+{ \left( ^{ n }{ { C }_{ 1 } } \right) }^{ 2 }+...{ \left( ^{ n }{ { C }_{ n } } \right) }^{ 2 }, where (nCi){ \left( ^{ n }{ { C }_{ i } } \right) } is the ithi^{th} coefficient of (1+x)n(1+x)^n expansion.
So, C12+C22+...+Cn2=(2nCn)=2n!n!n!C_1^2+C_2^2+...+C_n^2={ \left( ^{ 2n }{ { C }_{ n } } \right) }=\dfrac { 2n! }{ n!n! }

x^n এর সহগ নির্ণয় বিষয়ক টপিকের ওপরে পরীক্ষা দাও