দ্বিপদী বিস্তৃতি
(ax−ax)2n \left ( a x - \frac{a}{x} \right )^{2 n} (ax−xa)2n একটি দ্বিপদী রাশি।
a=1 হলে রাশিটির শেষ পদ কোনটি ?
1/x^(2n)
-1/x^(2n)
(-1/x)^(2n)
(-1)/(-x)^(2n)
a=1 a=1 a=1 হলে বিস্তৃতি হবে (x−1x)2n \left(x-\frac{1}{x}\right)^{2 n} (x−x1)2n ∴ \therefore ∴ শেষ পদ =(−1x)2n=1x2n =\left(\frac{-1}{x}\right)^{2 n}=\frac{1}{x^{2 n}} =(x−1)2n=x2n1
The first integral term in the expansion of (3+23)9(\sqrt{3}+\sqrt[3]{2})^{9}(3+32)9, is its
Let sn=1+q+q2+.................+qns_n = 1 + q + q^2 + ................. + q^nsn=1+q+q2+.................+qn and
Tn=1+(q+12)+(q+12)2+.........(q+12)nT_n = 1 + \left(\dfrac{q + 1}{2}\right) + \left(\dfrac{q + 1}{2}\right)^2 +......... \left(\dfrac{q + 1}{2}\right)^nTn=1+(2q+1)+(2q+1)2+.........(2q+1)n
where q is a real number and q ≠\neq= 1.
If 101C1+101C2.S1+......+101C101.S100=αT100,then α^{101}C_1 + ^{101}{C_2}.S_{1} +...... + ^{101}C_{101}.S_{100} = \alpha T_{100}, then\,\,\, \alpha101C1+101C2.S1+......+101C101.S100=αT100,thenα is equal to :-
The middle term in the expansion of (ax)n(a x)^n(ax)n is
The number of integral terms in expansion (32+58)256(\sqrt [2]{3} + \sqrt [ 8 ]{ 5 } )^{256}(23+85)256 is
