Q. 205.0( 1 Vote )

Prove that

To prove:

Formula used: (i)

(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn

(a+b)4 = 4C0a4 + 4C1a4-1b + 4C2a4-2b2 + 4C3a4-3b3 + 4C4b4

4C0a4 + 4C1a3b + 4C2a2b2 + 4C3a1b3 + 4C4b4 … (i)

(a-b)4 = 4C0a4 + 4C1a4-1(-b) + 4C2a4-2(-b)2 +4C3a4-3(-b)3+4C4(-b)4

4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4 … (ii)

(a+b)4 + (a-b)7 = [4C0a4 + 4C1a3b + 4C2a2b2 + 4C3a1b3 + 4C4b4] + [4C0a4 - 4C1a3b + 4C2a2b2 - 4C3ab3 + 4C4b4]

2[4C0a4 + 4C2a2b2 + 4C4b4]

2

2[(1)a4 + (6)a2b2 + (1)b4]

2[a4 + 6a2b2 + b4]

Therefore, (a+b)4 + (a-b)7 = 2[a4 + 6a2b2 + b4]

Now, putting a = 2 and b = in the above equation.

= 2[(2)4 + 6(2)2 ()2 + ()4]

= 2(16+24x+x2)

Hence proved.

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