এর বিস্তৃতিতে ধ্রুবক পদের মান নির্ণয় কর।
Created: 2 years ago |
Updated: 2 weeks ago
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(x^2-2+1/x^2)^6 এর বিস্তৃতিতে ধ্রুবক পদের মান নির্ণয়ের জন্য, আমরা নিম্নলিখিত পদগুলি গণনা করি:
T_0 = (1)^6 = 1
T_1 = 6(1)(-2) = -12
T_2 = (6)(6)(1)/2 = 18
T_3 = 6(12)(-2)/3! = -48
T_4 = (6)(6)(6)(1)/4! = 90
T_5 = 6(12)(48)/5! = -96
T_6 = (6)(6)(6)(6)(1)/6! = 36
T_7 = 6(12)(96)/7! = -324
T_8 = (6)(6)(6)(6)(6)(1)/8! = 54
T_9 = 6(12)(324)/9! = -756
T_{10} = (6)(6)(6)(6)(6)(6)(1)/10! = 36
T_{11} = 6(12)(756)/11! = -1296
T_{12} = (6)(6)(6)(6)(6)(6)(6)(1)/12! = 0
সুতরাং, (x^2-2+1/x^2)^6 এর বিস্তৃতিতে ধ্রুবক পদের মান হল,
T_0 + T_2 + T_4 + T_6 + T_8 + T_{10} + T_{12} = 1 + 18 + 90 + 36 + 54 + 36 + 0 = 924
অতএব, উত্তর হল 924।
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