Question
Question: If x be so small that its square and higher powers may be neglected, find the value of: \[\dfrac{{...
If x be so small that its square and higher powers may be neglected, find the value of:
(1+2x)51+(1−2x)51(1−53x)41+(1+65x)−6
Solution
Hint : As we have already the binomial expansion, so we will use this expansion in the question and then neglecting the highest power in it, we can easily find the solution of the given question. We will use the formula for binomial expansion, that is, (1+x)n=1+nx+2!n(n−1)+.........
Complete step-by-step answer :
Now from the question, we have to find the value of (1+2x)51+(1−2x)51(1−53x)41+(1+65x)−6
⇒(1+52x)+(1−10x)(1−53×41x)+(1+65×x(−6))
⇒2+103x2−20103x
⇒(2−20203x)21(1+203x)−1
⇒21(2−20103x)(1−203x)
⇒21(2−20109x)=1−40109x
⇒1−40109x , which is the required answer.
Note : Binomial Theorem
The theorem is that the method of expanding an expression which has been raised to any finite power. A theorem may be a powerful tool of expansion, which has application in Algebra, probability, etc. The theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theory , it's possible to expand the polynomial (x+y)n into a sum involving terms of the shape axbyc , where the exponents b and c are nonnegative integers with b + c = n, and therefore the coefficient a of every term may be a specific positive integer counting on n and b.
Important points to remember
1. The total number of terms in the expansion of (x+y)n are (n+1)
2. The sum of exponents of x and y is always n.
3. nC0, nC1, nC2, … .., nCn are called binomial coefficients and also represented by C0, C1, C2, ….., Cn
4. The binomial coefficients which are equidistant from the start and from the ending are equal i.e. nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.