দ্বিপদী বিস্তৃতি
(mx3-n/x2)15 একটি দ্বিপদী রাশি ।
3 তম পদের সহগ 105m13 হলে n এর মান কত ?
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৩য় পদ =15C2(mx3)15−2(−nx2)2=105 m13x39(−1)2n2x−4∴105 m13n2=105 m13 वा, n2=1∴n=±1 \begin{array}{l} \text { ৩য় পদ }={ }^{15} \mathrm{C}_{2}\left(\mathrm{mx}^{3}\right)^{15-2}\left(-\frac{\mathrm{n}}{\mathrm{x}^{2}}\right)^{2} \\ =105 \mathrm{~m}^{13} \mathrm{x}^{39}(-1)^{2} \mathrm{n}^{2} \mathrm{x}^{-4} \\ \therefore 105 \mathrm{~m}^{13} \mathrm{n}^{2}=105 \mathrm{~m}^{13} \text { वा, } \mathrm{n}^{2}=1 \quad \therefore \mathrm{n}= \pm 1\end{array} ৩য় পদ =15C2(mx3)15−2(−x2n)2=105 m13x39(−1)2n2x−4∴105 m13n2=105 m13 वा, n2=1∴n=±1
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
