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Solution :

`(1 + x)^101 (1 + x^2 - x)^100` <br>
`= (1 + x)(1 +x)^100 (1 + x^2 - x)^100` <br>
`= (1+x)[(1+x)(1-x+x^2)]^100` <br>
`= (1+x)(1+x^3)^100` <br>
`= (1+x)(a_o + a_1 x^3 + a_2x^9 + ..... + a_100 x^300)` <br>
`= a_0 + a_1 x^3 + a_2 x^6 + ..... + a_100 x^300` <br>
` a_0 x + a_1 x^4 + a_2 x^7 + .... + a_100 x^301` <br>
cant simplify so <br>
`101 xx 2 = 202` <br>
now,`2x^2 - 3x^3 - 5x` <br>
`5x^2 - 5x` <br>
Answer**Introduction**

**State & proof Binomial Theorem and Pascal triangle**

**Number of terms in expansion of following
(i) `(2x-3y)^9` (ii) `(2x+3y-4z)^n`**

**Expand `(x^2+2a)^5`; `(2x-3y)^4`;`(1+x+x^2)^3` by binomial theorem**

**(iii) Find an approximate value of `(0.99)^5` using the first three terms of its expansion.**

**Using binomial theorem prove that `(101)^50 gt(100)^50+(99)^50`**

**What is General and middle term in a binomial expansion**

**Term from the end of expansion**

**Coefficient of certain power of variable in binomial**

**finding term independent of variable**