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`sum_(i=0)^(n) sum_(j=0)^(n) (i + j) C_(i) C_(j) = sum_(i=0)^(n) sum_(j=0)^(n) iC_(i) C_(j) + sum_(i=0)^(n) sum_(j=0)^(n) jC_(i) C_(j)` <br> `=sum_(i=0)^(n)i C_(i) (sum_(j=0)^(n) C_(j)) + sum_(j=0)^(n) jC_(j) (sum_(i=0)^(n)C_(i))` <br> `= sum_(i=0)^(n)i C_(i)(2^(n)) + sum_(i=0)^(n)j C_(j)(2^(n))` <br> ` = 2^(n) sum_(i=0)^(n)i ""^(n)C_(i) + 2^(n) sum_(j=0)^(n) j""^(n)C_(j) ` <br> `=2^(n) sum_(i=0)^(n)i.(n)/(i) .""^(n-1)C_(i-1)+ 2^(n) sum_(j=0)^(n)j * (n)/(j) * ""^(n-1)C_(j-1)` <br> `=n.2^(n) sum_(i=0)^(n) .""^(n-1)C_(i-1)+ n.2^(n) sum_(j=0)^(n) * ""^(n-1)C_(j-1)` <br> `= n.2^(n).2^(n-1) + n.2^(n) * 2^(n-1)` <br> ` = n*2*2^(2n-1) = n*2^(2n)` .Transcript

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00:00 - 00:59 | Vivan about today's question as if oneplus x whole to the power n is equal to 30 + 7 x + 2 x square up to see and then 202 and submission is equal to zero to one Eye Plus Jaisi icj so we can separate at so weak at summation is equals to zero to an emissions is equal to 0 to n i c i c plus submission is equal to zero to and summation day is equal to zero to an jci so if we can buy it is it isn't I'm submission is equal to zero to one ICI summation Jay is equal to zero to ncj + summation he is equal to zero to and jcj summation is equals to zero to MCI |

01:00 - 01:59 | so that's some of this series and the sum of this series is equals to 2 to the power and so we can take the common so with pictures 2 to the power and submission is equal to zero to an ICI plus summation Jaise = 202 n j j n v can Rai we can know that NCR is equals to 1 upon our end n -1 Siya -1 so we can write it as 2 to the power and submission is equal to zero to an end to end up on i n -1 -1 + summation Jaise = 202 and J into an app on n -1 |

02:00 - 02:59 | -1 this chair is Desai and decide is cancel out and from both the term we can have common riches and 2 to the power and n in bracket summation ise equal to 0 to N N -1 -1 + submission there is equals to Zero 2 N N N -1 c.j. -1 so the sum of the series and the sum of the series if there is also the sum of the series 2 to the power and but in place of an address and -1 so the sum of the series as to the power and into n 2 to the power n minus 1 + 2 to the power and -5 so we can take write a Tier 2 to the power |

03:00 - 03:59 | to an end to end it is our final answer thank you |

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**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**