Find x + 1 6 + x - 1 6 . Hence or otherwise evaluate 2 + 1 6 + 2 - 1 6 .

To find: x + 1 6 + x - 1 6 We know that, Binomial expansion formula is given by: ( a + b ) n = n C 0 a n + n C 1 a n - 1 b + n C 2 a n - 2 b 2 + … + n C n - 1 a b n - 1 + n C n b n By using binomial expansion, ( x + 1 ) 6 = C 0 6 x 6 + C 1 6 x 5 + C 2 6 x 4 + C 3 6 x 3 + C 4 6 x 2 + C 5 6 x + C 6 6 x 0 and ( x - 1 ) 6 = C 0 6 x 6 - C 1 6 x 5 + C 2 6 x 4 - C 3 6 x 3 + C 4 6 x 2 - C 5 6 x + C 6 6 x 0 ∴ ( x + 1 ) 6 + ( x - 1 ) 6 = 2 C 0 6 x 6 + C 2 6 x 4 + C 4 6 x 2 + C 6 6 = 2 x 6 + 15 x 4 + 15 x 2 + 1 … . ( i ) In equation ( i ) we put x = 2 , 2 + 1 6 + 2 - 1 6 = 2 2 6 + 15 2 4 + 15 2 2 + 1 = 2 8 + 60 + 30 + 1 = 2 99 = 198

Find x + 1 6 + x - 1 6 . Hence or otherwise evaluate 2 + 1 6 + 2 - 1 6 .

To find: x + 1 6 + x - 1 6 We know that, Binomial expansion formula is given by: ( a + b ) n = n C 0 a n + n C 1 a n - 1 b + n C 2 a n - 2 b 2 + … + n C n - 1 a b n - 1 + n C n b n By using binomial expansion, ( x + 1 ) 6 = C 0 6 x 6 + C 1 6 x 5 + C 2 6 x 4 + C 3 6 x 3 + C 4 6 x 2 + C 5 6 x + C 6 6 x 0 and ( x - 1 ) 6 = C 0 6 x 6 - C 1 6 x 5 + C 2 6 x 4 - C 3 6 x 3 + C 4 6 x 2 - C 5 6 x + C 6 6 x 0 ∴ ( x + 1 ) 6 + ( x - 1 ) 6 = 2 C 0 6 x 6 + C 2 6 x 4 + C 4 6 x 2 + C 6 6 = 2 x 6 + 15 x 4 + 15 x 2 + 1 … . ( i ) In equation ( i ) we put x = 2 , 2 + 1 6 + 2 - 1 6 = 2 2 6 + 15 2 4 + 15 2 2 + 1 = 2 8 + 60 + 30 + 1 = 2 99 = 198

E M B I B E

Mathematics Textbook for Class 11>Chapter -1 - Binomial Theorem>EXERCISE 8.1>Q 12

HARD

11th CBSE

IMPORTANT

Earn 100

Find ${(x + 1)}^{6} + {(x - 1)}^{6} .$ Hence or otherwise evaluate ${(\sqrt{2} + 1)}^{6} + {(\sqrt{2} - 1)}^{6} .$

Important Points to Remember in Chapter -1 - Binomial Theorem from NCERT Mathematics Textbook for Class 11 Solutions

1. Binomial Expression:

An expression consisting of two terms, connected by $+$ or $-$ sign is called binomial expression.

2. Binomial Theorem for Positive Integer:

If $x$ and $a$ are real numbers, then for all $n \in N,$ we have ${(x + a)}^{n} =^{n} C_{0} x^{n} a^{0} +^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} + ..... +^{n} C_{r} x^{n - r} a^{r} + . ... +^{n} C_{n} x^{0} a^{n}$ i.e, ${(x + a)}^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{n - r} a^{r}$ .

3. Properties of Binomial Theorem for Positive Integer:

(i) It has $(n + 1)$ terms.

(ii) The sum of the indices of $x$ and $= 2 [x^{6} + 15 x^{4} + 15 x^{2} + 1] \dots . (i)$ in each term is $n .$

(iii) The coefficients of terms equidistant from the beginning and the end are equal.

4. General term:

The general term in the expansion of $(x + a)^{n}$ is $T_{r + 1} =^{n} C_{r} x^{n - r} a^{r}$

(i) Replacing $a$ by $- a$ in the expansion of ${(x + a)}^{n}$ we get ${(x - a)}^{n} = {}^{n}C_{0} x^{n} a^{0} -^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} + . ... + {(- 1)}^{r}^{n} C_{r} x^{n - r} a^{r} + ..... + {(- 1)}^{n}^{n} C_{n} x^{0} a^{n}$

(ii) The general term in the expansion of ${(x - a)}^{n}$ is $T_{r + 1} = {(- 1)}^{r}^{n} C_{r} x^{n - r} a^{r}$

(iii) Putting $x = 1$ and replacing $a$ by $x$ in the expansion of ${(x + a)}^{n},$ we get ${(1 + x)}^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n} = \sum_{r = 0}^{n}^{n} C_{r} x^{r}$

This is the expansion of ${(1 + x)}^{n}$ in ascending powers of $x .$ In this case, $T_{r + 1} =^{n} C_{r} x^{r}$

(iv) Putting $a = 1$ in the expansion of ${(x + a)}^{n},$ we get ${(x + 1)}^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} +^{n} C_{2} x^{n - 2} + \dots +^{n} C_{n} x^{0} = \sum_{r = 0}^{n}^{n} C_{r} x^{n - r}$

This is the expansion of ${(x + 1)}^{n}$ in descending powers of $x .$ In this case, $T_{r + 1} =^{n} C_{r} x^{n - r}$

(v) ${(x + a)}^{n} + {(x - a)}^{n} = 2 \{^{n} C_{0} x^{n} a^{0} +^{n} C_{2} x^{n - 2} a^{2} + . ...\}$

$= 2$ (Sum of the odd terms in the expansion of ${(x + a)}^{n}$ )

(vi) ${(x + a)}^{n} - {(x - a)}^{n} = 2 \{^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{3} x^{n - 3} a^{3} + . ...\}$

$= 2$ (Sum of the even terms in the expansion of ${(x + a)}^{n}$ )

(vii) If $n$ is odd, then $\{{(x + a)}^{n} + {(x - a)}^{n}\}$ and $\{{(x + a)}^{n} - {(x - a)}^{n}\}$ both have $(\frac{n + 1}{2})$ terms.

(viii) If $n$ is even, then $\{{(x + a)}^{n} + {(x - a)}^{n}\}$ has $(\frac{n}{2} + 1)$ terms where as $\{{(x + a)}^{n} - {(x - a)}^{n}\}$ has $(\frac{n}{2})$ terms.