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Question

Mathematics Question on Binomial theorem

Let n ≥ 5 be an integer. If 9n – 8n – 1 = 64α and 6n – 5n – 1 = 25β, then α – β is equal to

A

1+nC2(8−5)+nC3(82−52)+...+nCn(8n−1−5n−1)1 + ^nC_2 (8-5) + ^nC_3(8² - 5²) + ... + ^nC_n ( 8^{n-1} - 5^{n-1} )

B

1+nC3(8−5)+nC4(82−52)+...+nCn(8n−2\-5n−2)1 + ^nC_3 (8-5) + ^nC_4(8² - 5²) + ... + ^nC_n ( 8^{n-2} \- 5^{n-2} )

C

nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2)^nC_3 (8-5) + ^nC_4 ( 8²-5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )

D

nC4(8−5)+nC5(82−52)+...+nCn(8n−3\-5n−3)^nC_4 (8-5) + ^nC_5 ( 8²-5²) + ... + ^nC_n ( 8^{n-3} \- 5^{n-3} )

Answer

nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2)^nC_3 (8-5) + ^nC_4 ( 8²-5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )

Explanation

Solution

The correct answer is (C) : nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2)^nC_3 (8-5) + ^nC_4 ( 8²-5²) + ... + ^nC_n ( 8^{n-2} - 5^{n-2} )
(1 + 8)n _– _8n – 1 = 64α
⇒1+8n+nC282+nC383+.....+nCn8n−8n−1=64α⇒ 1 + 8n + ^nC_28^2 + ^nC_38^3 + ..... +^nC_n8^n - 8n-1 = 64α
⇒α=nC2+nC38+nC482+.....+nCn8n−2......(i)⇒ α = ^nC_2 + ^nC_38 + ^nC_48^2 + ..... + ^nC_n8^{n-2} ...... (i)
Similarly
(1+5)n - 5n-1=25β
⇒1+5n+nC252+nC353+.....+nCn5n−5n−1=25β⇒ 1 + 5n + ^nC_25^2 + ^nC_35^3 + ..... + ^nC_n5^n - 5n - 1 = 25β
⇒β=nC2+nC3.5+nC4.52+......+nCn5n−2.....(ii)⇒ β = ^nC_2 + ^nC_3.5 + ^nC_4.5^2 + ...... + ^nC_n 5^{n-2} ..... (ii)
α−β=nC3(8−5)+nC4(82−52)+.....+nCn(8n−2−5n−2)α - β = ^nC_3(8-5) + ^nC_4 (8^2-5^2) + ..... + ^nC_n(8^{n-2} - 5^{n-2})