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EXERCISE 4.3 · Q38

Q.In the following expansion, find the indicated term: (3a+4a)13\left(3a+\dfrac{4}{a}\right)^{13}, 10th term.

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Here abase=3a,b=4a,n=13a_{\text{base}}=3a,b=\dfrac4a,n=13. For t10t_{10}, r=9r=9. t10=13C9(3a)4(4a)9=715⋅81a4⋅262144a9=715×81×262144×a−5t_{10}={}^{13}C_9(3a)^4\left(\dfrac4a\right)^9=715\cdot81a^4\cdot\dfrac{262144}{a^9}=715\times81\times262144\times a^{-5}. Now 81×262144=2123366481\times262144=21233664, …

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