RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.6

Question 1. Find the sum of the series: \( 1 + \frac{2}{3} \cdot \frac{1}{2} + \frac{2 \cdot 5}{3 \cdot 6} \cdot \left(\frac{1}{2}\right)^2 + \frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \left(\frac{1}{2}\right)^3 + \dots \)
Answer:
Let the given series be compared with the binomial expansion \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \)
Comparing the second term:
\( nx = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} \)
Squaring both sides:
\( n^2x^2 = \frac{1}{9} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n-1)}{2!}x^2 = \frac{2 \cdot 5}{3 \cdot 6} \cdot \left(\frac{1}{2}\right)^2 \)
\( \frac{n(n-1)}{2}x^2 = \frac{10}{18} \cdot \frac{1}{4} = \frac{5}{9} \cdot \frac{1}{4} = \frac{5}{36} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n-1)x^2 / 2}{n^2x^2} = \frac{5/36}{1/9} \)
\( \frac{n-1}{2n} = \frac{5}{36} \cdot 9 \)
\( \frac{n-1}{2n} = \frac{5}{4} \)
Multiply both sides by \( 4n \):
\( 4(n-1) = 10n \)
\( 4n - 4 = 10n \)
\( -4 = 10n - 4n \)
\( -4 = 6n \)
\( n = -\frac{4}{6} \)
\( n = -\frac{2}{3} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{3} \)):
\( (-\frac{2}{3})x = \frac{1}{3} \)
Multiply both sides by \( 3 \):
\( -2x = 1 \)
\( x = -\frac{1}{2} \)
Therefore, the sum of the series is \( (1+x)^n \):
\( \text{Sum} = \left(1 + \left(-\frac{1}{2}\right)\right)^{-2/3} \)
\( = \left(1 - \frac{1}{2}\right)^{-2/3} \)
\( = \left(\frac{1}{2}\right)^{-2/3} \)
\( = (2)^{-(-2/3)} \)
\( = (2)^{2/3} \)
\( = (2^2)^{1/3} \)
\( = (4)^{1/3} \)
In simple words: We matched the given series with a standard mathematical expansion formula. By comparing its terms, we found the 'n' and 'x' values. Then, we put these values back into the formula to calculate the final sum, which came out to be the cube root of 4.

🎯 Exam Tip: When dealing with binomial series, carefully compare the coefficients of the terms to find 'n' and 'x'. Ensure you correctly handle fractions and negative exponents during calculations.

Question 2. Find the sum of the series: \( 1 + \frac{1}{3} \cdot \frac{1}{4} + \frac{1 \cdot 4}{3 \cdot 6} \cdot \left(\frac{1}{4}\right)^2 + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9} \cdot \left(\frac{1}{4}\right)^3 + \dots \)
Answer:
Let the given series be compared with the binomial expansion \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \)
Comparing the second term:
\( nx = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n-1)}{2!}x^2 = \frac{1 \cdot 4}{3 \cdot 6} \cdot \left(\frac{1}{4}\right)^2 \)
\( \frac{n(n-1)}{2}x^2 = \frac{4}{18} \cdot \frac{1}{16} = \frac{2}{9} \cdot \frac{1}{16} = \frac{2}{144} = \frac{1}{72} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n-1)x^2 / 2}{n^2x^2} = \frac{1/72}{1/144} \)
\( \frac{n-1}{2n} = \frac{1}{72} \cdot 144 \)
\( \frac{n-1}{2n} = 2 \)
Multiply both sides by \( 2n \):
\( n-1 = 4n \)
\( -1 = 4n - n \)
\( -1 = 3n \)
\( n = -\frac{1}{3} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{12} \)):
\( \left(-\frac{1}{3}\right)x = \frac{1}{12} \)
Multiply both sides by \( 3 \):
\( -x = \frac{3}{12} \)
\( -x = \frac{1}{4} \)
\( x = -\frac{1}{4} \)
Therefore, the sum of the series is \( (1+x)^n \):
\( \text{Sum} = \left(1 + \left(-\frac{1}{4}\right)\right)^{-1/3} \)
\( = \left(1 - \frac{1}{4}\right)^{-1/3} \)
\( = \left(\frac{3}{4}\right)^{-1/3} \)
\( = \left(\frac{4}{3}\right)^{1/3} \)
In simple words: We used the binomial expansion formula to break down the given series. By matching the parts, we found the values for 'n' and 'x'. Putting these back into the formula gave us the sum, which is the cube root of four-thirds.

🎯 Exam Tip: Remember that binomial series are often of the form \( (1+x)^n \) or \( (1-x)^n \). The sign of 'x' in your calculated series determines if it's an alternating sum or not.

Question 3. Find the sum of the series: \( 1 + \frac{1}{4} + \frac{1 \cdot 4}{4 \cdot 8} + \frac{1 \cdot 4 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \)
Answer:
Let the given series be compared with the binomial expansion \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \)
Comparing the second term:
\( nx = \frac{1}{4} \)
Squaring both sides:
\( n^2x^2 = \frac{1}{16} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n-1)}{2!}x^2 = \frac{1 \cdot 4}{4 \cdot 8} \)
\( \frac{n(n-1)}{2}x^2 = \frac{4}{32} = \frac{1}{8} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n-1)x^2 / 2}{n^2x^2} = \frac{1/8}{1/16} \)
\( \frac{n-1}{2n} = \frac{1}{8} \cdot 16 \)
\( \frac{n-1}{2n} = 2 \)
Multiply both sides by \( 2n \):
\( n-1 = 4n \)
\( -1 = 4n - n \)
\( -1 = 3n \)
\( n = -\frac{1}{3} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{4} \)):
\( \left(-\frac{1}{3}\right)x = \frac{1}{4} \)
Multiply both sides by \( 3 \):
\( -x = \frac{3}{4} \)
\( x = -\frac{3}{4} \)
Therefore, the sum of the series is \( (1+x)^n \):
\( \text{Sum} = \left(1 + \left(-\frac{3}{4}\right)\right)^{-1/3} \)
\( = \left(1 - \frac{3}{4}\right)^{-1/3} \)
\( = \left(\frac{1}{4}\right)^{-1/3} \)
\( = (4)^{1/3} \)
In simple words: We matched the given series with a common binomial expansion. After finding 'n' and 'x' by comparing the terms, we calculated the total sum using the formula. The final answer is the cube root of 4.

🎯 Exam Tip: Practice identifying the pattern of binomial series expansions. The form of the coefficients (e.g., \( 1 \cdot 4 \cdot 7 \dots \)) helps determine the value of 'n'.

Question 4. Find the sum of the series: \( 1 + \frac{1}{10} + \frac{1 \cdot 4}{10 \cdot 20} + \frac{1 \cdot 4 \cdot 7}{10 \cdot 20 \cdot 30} + \dots \)
Answer:
Let the given series be compared with the binomial expansion \( (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \)
Comparing the second term:
\( nx = \frac{1}{10} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n-1)}{2!}x^2 = \frac{1 \cdot 4}{10 \cdot 20} \)
\( \frac{n(n-1)}{2}x^2 = \frac{4}{200} = \frac{1}{50} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n-1)x^2 / 2}{n^2x^2} = \frac{1/50}{1/100} \)
\( \frac{n-1}{2n} = \frac{1}{50} \cdot 100 \)
\( \frac{n-1}{2n} = 2 \)
Multiply both sides by \( 2n \):
\( n-1 = 4n \)
\( -1 = 4n - n \)
\( -1 = 3n \)
\( n = -\frac{1}{3} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{10} \)):
\( \left(-\frac{1}{3}\right)x = \frac{1}{10} \)
Multiply both sides by \( 3 \):
\( -x = \frac{3}{10} \)
\( x = -\frac{3}{10} \)
Therefore, the sum of the series is \( (1+x)^n \):
\( \text{Sum} = \left(1 + \left(-\frac{3}{10}\right)\right)^{-1/3} \)
\( = \left(1 - \frac{3}{10}\right)^{-1/3} \)
\( = \left(\frac{7}{10}\right)^{-1/3} \)
\( = \left(\frac{10}{7}\right)^{1/3} \)
In simple words: We identified the given series as a binomial expansion. By matching its terms to the formula, we found the values for 'n' and 'x'. Then, we used these values to compute the total sum, which is the cube root of ten-sevenths.

🎯 Exam Tip: Be careful with signs when substituting negative values for 'x'. A series with all positive terms can still result from a negative 'x' if 'n' is also negative and the overall expansion results in positive terms.

Question 5. Find the sum of the series: \( 1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1 \cdot 3}{2 \cdot 4} \cdot \left(\frac{1}{2}\right)^2 - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \left(\frac{1}{2}\right)^3 + \dots \)
Answer:
Let the given series be compared with the binomial expansion \( (1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots \)
Comparing the second term:
\( ny = -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4} \)
Squaring both sides:
\( n^2y^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n-1)}{2!}y^2 = \frac{1 \cdot 3}{2 \cdot 4} \cdot \left(\frac{1}{2}\right)^2 \)
\( \frac{n(n-1)}{2}y^2 = \frac{3}{8} \cdot \frac{1}{4} = \frac{3}{32} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n-1)y^2 / 2}{n^2y^2} = \frac{3/32}{1/16} \)
\( \frac{n-1}{2n} = \frac{3}{32} \cdot 16 \)
\( \frac{n-1}{2n} = \frac{3}{2} \)
Multiply both sides by \( 2n \):
\( n-1 = 3n \)
\( -1 = 3n - n \)
\( -1 = 2n \)
\( n = -\frac{1}{2} \)
Substitute the value of \( n \) into equation (i) (\( ny = -\frac{1}{4} \)):
\( \left(-\frac{1}{2}\right)y = -\frac{1}{4} \)
Multiply both sides by \( -2 \):
\( y = (-\frac{1}{4}) \cdot (-2) \)
\( y = \frac{1}{2} \)
Therefore, the sum of the series is \( (1+y)^n \):
\( \text{Sum} = \left(1 + \frac{1}{2}\right)^{-1/2} \)
\( = \left(\frac{3}{2}\right)^{-1/2} \)
\( = \left(\frac{2}{3}\right)^{1/2} \)
\( = \sqrt{\frac{2}{3}} \)
In simple words: We used the binomial expansion to find the power 'n' and the variable 'y' that create this alternating series. After finding 'n' as -1/2 and 'y' as 1/2, we plugged them back into the formula to get the final sum, which is the square root of two-thirds.

🎯 Exam Tip: When the terms of a series alternate in sign, it often indicates a negative value for the variable 'x' (or 'y') in the binomial expansion \( (1+x)^n \).

Question 6. Prove that: \( \sqrt{2} = 1 + \frac{1}{2^2} + \frac{1 \cdot 3}{2! 2^4} + \frac{1 \cdot 3 \cdot 5}{3! 2^6} + \dots \)
Answer:
Let the given series be denoted by \( S \). We compare it with the expansion of \( (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots \)
The series can be written as:
\( S = 1 + \frac{1}{4} + \frac{1 \cdot 3}{2 \cdot 16} + \frac{1 \cdot 3 \cdot 5}{6 \cdot 64} + \dots \)
\( S = 1 + \frac{1}{4} + \frac{3}{32} + \frac{15}{384} + \dots \)
Comparing the second term:
\( nx = \frac{1}{4} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n+1)}{2!}x^2 = \frac{1 \cdot 3}{2! 2^4} \)
\( \frac{n(n+1)}{2}x^2 = \frac{3}{2 \cdot 16} = \frac{3}{32} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n+1)x^2 / 2}{n^2x^2} = \frac{3/32}{1/16} \)
\( \frac{n+1}{2n} = \frac{3}{32} \cdot 16 \)
\( \frac{n+1}{2n} = \frac{3}{2} \)
Multiply both sides by \( 2n \):
\( n+1 = 3n \)
\( 1 = 3n - n \)
\( 1 = 2n \)
\( n = \frac{1}{2} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{4} \)):
\( \left(\frac{1}{2}\right)x = \frac{1}{4} \)
Multiply both sides by \( 2 \):
\( x = \frac{2}{4} \)
\( x = \frac{1}{2} \)
Therefore, the sum of the series is \( (1-x)^{-n} \):
\( S = \left(1 - \frac{1}{2}\right)^{-1/2} \)
\( = \left(\frac{1}{2}\right)^{-1/2} \)
\( = (2)^{1/2} \)
\( = \sqrt{2} \)
Hence Proved.
In simple words: We took the right side of the equation and treated it as a binomial expansion. By comparing its terms to the general form for \( (1-x)^{-n} \), we found that 'n' is 1/2 and 'x' is 1/2. Plugging these values back into the formula gave us the sum, which is exactly \( \sqrt{2} \), proving the statement.

🎯 Exam Tip: When proving an identity involving series, first identify the type of binomial expansion (e.g., \( (1+x)^n \) or \( (1-x)^{-n} \)), then find 'n' and 'x' from the series terms. Finally, substitute these values to calculate the sum.

Question 7. Prove that: \( \frac{5\sqrt{2}}{7} = \left[1 + \frac{1}{10^2} + \frac{1 \cdot 3}{1 \cdot 2 \cdot 10^4} + \frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 10^6} + \dots\right] \)
Answer:
Let the series on the Right Hand Side (RHS) be \( S \). We compare it with the expansion of \( (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots \)
The series can be written as:
\( S = 1 + \frac{1}{100} + \frac{1 \cdot 3}{2 \cdot 10000} + \frac{1 \cdot 3 \cdot 5}{6 \cdot 1000000} + \dots \)
Comparing the second term:
\( nx = \frac{1}{100} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{100}\right)^2 = \frac{1}{10000} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n+1)}{2!}x^2 = \frac{1 \cdot 3}{1 \cdot 2 \cdot 10^4} \)
\( \frac{n(n+1)}{2}x^2 = \frac{3}{2 \cdot 10000} = \frac{3}{20000} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n+1)x^2 / 2}{n^2x^2} = \frac{3/20000}{1/10000} \)
\( \frac{n+1}{2n} = \frac{3}{20000} \cdot 10000 \)
\( \frac{n+1}{2n} = \frac{3}{2} \)
Multiply both sides by \( 2n \):
\( n+1 = 3n \)
\( 1 = 3n - n \)
\( 1 = 2n \)
\( n = \frac{1}{2} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{100} \)):
\( \left(\frac{1}{2}\right)x = \frac{1}{100} \)
Multiply both sides by \( 2 \):
\( x = \frac{2}{100} \)
\( x = \frac{1}{50} \)
Therefore, the sum of the series is \( (1-x)^{-n} \):
\( S = \left(1 - \frac{1}{50}\right)^{-1/2} \)
\( = \left(\frac{49}{50}\right)^{-1/2} \)
\( = \left(\frac{50}{49}\right)^{1/2} \)
\( = \frac{\sqrt{50}}{\sqrt{49}} \)
\( = \frac{\sqrt{25 \cdot 2}}{7} \)
\( = \frac{5\sqrt{2}}{7} \)
Hence Proved.
In simple words: We analyzed the series on the right side and recognized it as a binomial expansion of the form \( (1-x)^{-n} \). By comparing its terms, we found that 'n' is 1/2 and 'x' is 1/50. When we put these values into the formula, the sum calculated was \( \frac{5\sqrt{2}}{7} \), which proves the given statement.

🎯 Exam Tip: For "prove that" questions, always start with one side (usually the series) and simplify it to match the other side. Clearly showing the steps for finding 'n' and 'x' is crucial.

Question 8. Prove that: \( \left(\frac{3}{2}\right)^{1/3} = 1 + \frac{1}{3^2} + \frac{1 \cdot 4}{1 \cdot 2 \cdot 3^4} + \frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 3^6} + \dots \)
Answer:
Let the series on the Right Hand Side (RHS) be \( S \). We compare it with the expansion of \( (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots \)
The series can be written as:
\( S = 1 + \frac{1}{9} + \frac{1 \cdot 4}{2 \cdot 81} + \frac{1 \cdot 4 \cdot 7}{6 \cdot 729} + \dots \)
Comparing the second term:
\( nx = \frac{1}{9} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{9}\right)^2 = \frac{1}{81} \quad \dots(\text{i}) \)
Comparing the third term:
\( \frac{n(n+1)}{2!}x^2 = \frac{1 \cdot 4}{1 \cdot 2 \cdot 3^4} \)
\( \frac{n(n+1)}{2}x^2 = \frac{4}{2 \cdot 81} = \frac{2}{81} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n+1)x^2 / 2}{n^2x^2} = \frac{2/81}{1/81} \)
\( \frac{n+1}{2n} = 2 \)
Multiply both sides by \( 2n \):
\( n+1 = 4n \)
\( 1 = 4n - n \)
\( 1 = 3n \)
\( n = \frac{1}{3} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{9} \)):
\( \left(\frac{1}{3}\right)x = \frac{1}{9} \)
Multiply both sides by \( 3 \):
\( x = \frac{3}{9} \)
\( x = \frac{1}{3} \)
Therefore, the sum of the series is \( (1-x)^{-n} \):
\( S = \left(1 - \frac{1}{3}\right)^{-1/3} \)
\( = \left(\frac{2}{3}\right)^{-1/3} \)
\( = \left(\frac{3}{2}\right)^{1/3} \)
Hence Proved.
In simple words: We recognized the given series as a binomial expansion. By comparing the terms of the series with the formula for \( (1-x)^{-n} \), we found the values for 'n' and 'x'. Substituting these values back into the formula gave us the sum \( \left(\frac{3}{2}\right)^{1/3} \), which proves the statement.

🎯 Exam Tip: When the coefficients in the numerator are consecutive odd numbers (1, 1⋅3, 1⋅3⋅5...), it often suggests that 'n' might be a fraction like 1/2 or 1/3, especially for negative exponents.

Question 9. If \( y = \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots \), then prove that \( y^2 + 2y - 2 = 0 \).
Answer:
Let the given series be \( y \). We can write it as:
\( y = \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots \)
This series is of the form \( (1-x)^{-n} - 1 = nx + \frac{n(n+1)}{2!}x^2 + \dots \)
Let \( S = 1+y = 1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots \)
Comparing \( S \) with \( (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots \)
Comparing the second term of \( S \):
\( nx = \frac{1}{3} \)
Squaring both sides:
\( n^2x^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \quad \dots(\text{i}) \)
Comparing the third term of \( S \):
\( \frac{n(n+1)}{2!}x^2 = \frac{1 \cdot 3}{3 \cdot 6} \)
\( \frac{n(n+1)}{2}x^2 = \frac{3}{18} = \frac{1}{6} \quad \dots(\text{ii}) \)
Now, divide equation (ii) by equation (i):
\( \frac{n(n+1)x^2 / 2}{n^2x^2} = \frac{1/6}{1/9} \)
\( \frac{n+1}{2n} = \frac{1}{6} \cdot 9 \)
\( \frac{n+1}{2n} = \frac{3}{2} \)
Multiply both sides by \( 2n \):
\( n+1 = 3n \)
\( 1 = 3n - n \)
\( 1 = 2n \)
\( n = \frac{1}{2} \)
Substitute the value of \( n \) into equation (i) (\( nx = \frac{1}{3} \)):
\( \left(\frac{1}{2}\right)x = \frac{1}{3} \)
Multiply both sides by \( 2 \):
\( x = \frac{2}{3} \)
Therefore, the sum of the series \( S = 1+y \) is \( (1-x)^{-n} \):
\( S = \left(1 - \frac{2}{3}\right)^{-1/2} \)
\( = \left(\frac{1}{3}\right)^{-1/2} \)
\( = (3)^{1/2} \)
\( S = \sqrt{3} \)
Since \( S = 1+y \), we have \( 1+y = \sqrt{3} \)
\( y = \sqrt{3} - 1 \)
Now, we need to prove \( y^2 + 2y - 2 = 0 \). Substitute the value of \( y \):
\( (\sqrt{3} - 1)^2 + 2(\sqrt{3} - 1) - 2 \)
\( = (\sqrt{3})^2 - 2\sqrt{3}(1) + (1)^2 + 2\sqrt{3} - 2 - 2 \)
\( = 3 - 2\sqrt{3} + 1 + 2\sqrt{3} - 2 - 2 \)
\( = (3 + 1 - 2 - 2) + (-2\sqrt{3} + 2\sqrt{3}) \)
\( = 0 + 0 \)
\( = 0 \)
Hence Proved.
In simple words: We added 1 to the given series to make it match a standard binomial expansion. By comparing the terms, we found the 'n' and 'x' values, which allowed us to calculate the value of \( 1+y \) as \( \sqrt{3} \). This means \( y \) is \( \sqrt{3}-1 \). Finally, we plugged this value of \( y \) into the equation \( y^2+2y-2=0 \) and showed that it equals zero, proving the statement.

🎯 Exam Tip: If a series starts from the second term of a binomial expansion, consider adding the missing first term (usually 1) to form a complete expansion, then subtract it at the end to find the sum of the original series.

Question 10. Prove that \( (1+x)^n = 2^n \left[1 - \frac{n(1-x)}{1+x} + \frac{n(n+1)}{2!} \left(\frac{1-x}{1+x}\right)^2 - \dots\right] \)
Answer:
Let's consider the Right Hand Side (RHS) of the equation:
\( \text{RHS} = 2^n \left[1 - \frac{n(1-x)}{1+x} + \frac{n(n+1)}{2!} \left(\frac{1-x}{1+x}\right)^2 - \dots\right] \)
Let \( y = \frac{1-x}{1+x} \).
Then the expression inside the square brackets becomes:
\( \left[1 - ny + \frac{n(n+1)}{2!}y^2 - \dots\right] \)
This is the binomial expansion of \( (1+y)^{-n} \).
Therefore, the RHS can be written as:
\( \text{RHS} = 2^n (1+y)^{-n} \)
Substitute back \( y = \frac{1-x}{1+x} \):
\( \text{RHS} = 2^n \left(1 + \frac{1-x}{1+x}\right)^{-n} \)
Combine the terms inside the parenthesis:
\( \text{RHS} = 2^n \left(\frac{1+x + 1-x}{1+x}\right)^{-n} \)
\( \text{RHS} = 2^n \left(\frac{2}{1+x}\right)^{-n} \)
Apply the negative exponent:
\( \text{RHS} = 2^n \left(\frac{1+x}{2}\right)^{n} \)
Separate the terms:
\( \text{RHS} = 2^n \cdot \frac{(1+x)^n}{2^n} \)
Cancel out \( 2^n \):
\( \text{RHS} = (1+x)^n \)
This is equal to the Left Hand Side (LHS).
Hence Proved.
In simple words: We started with the complicated right side of the equation. We noticed that the part inside the square brackets looked like a binomial expansion. By setting \( y = \frac{1-x}{1+x} \), we recognized it as the expansion of \( (1+y)^{-n} \). Then, we replaced \( y \) with its original expression, simplified the terms, and found that the entire right side simplifies to \( (1+x)^n \), which is exactly the left side of the equation.

🎯 Exam Tip: For complex "prove that" questions involving binomial series, look for a substitution that simplifies the series into a recognizable standard binomial expansion form. Algebraic manipulation is key to matching the LHS and RHS.