CANCELLATION
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1.
Sol.

Proof
𝑛
∑𝑛𝑘=0(𝑛𝑘)(𝑛−𝑘
) = (2𝑛
)
𝑛

Consider the expression (𝑥 + 𝑦)2𝑛
The coefficient of 𝑥 𝑛 𝑦 𝑛 in the expansion can be calculated as follows
Method 1
Using the binomial theorem directly,
we get the coefficient of 𝑥 𝑛 𝑦 𝑛 in the expansion of (𝑥 + 𝑦)2𝑛 to be (2𝑛
).
𝑛
Method 2
Rewriting the expression (𝑥 + 𝑦)2𝑛 𝑎𝑠 (𝑥 + 𝑦)𝑛 (𝑥 + 𝑦)𝑛
To obtain the 𝑥 𝑛 𝑦 𝑛 in the product,
Consider the term with 𝑥 𝑘 𝑦 𝑛−𝑘 in the first expansion. It should be multiplied with corresponding term
𝑥 𝑛−𝑘 𝑦 𝑘 in the second expansion for every value of k from 0 to n
From the binomial theorem,
we know that, in the expansion of (𝑥 + 𝑦)𝑛 , the coefficient of 𝑥 𝑛−𝑘 𝑦 𝑘 will be (𝑛𝑘).
Similarly, , in the expansion of (𝑥 + 𝑦)𝑛 , the coefficient of 𝑥 𝑛−𝑘 𝑦 𝑘 will be (𝑛𝑘).
𝑛
Therefore, multiplying the corresponding coefficients, we get (𝑛𝑘)(𝑛−𝑘
) as the coefficient of 𝑥 𝑛 𝑦 𝑛 from
the term 𝑥 𝑘 𝑦 𝑛−𝑘 .

And this should be calculated for all the values of k.
Therefore, coefficient of the term 𝑥 𝑛 𝑦 𝑛 in the product (𝑥 + 𝑦)𝑛 (𝑥 + 𝑦)𝑛 can be written as the sum of
all the corresponding coefficients which give rise to the term 𝑥 𝑛 𝑦 𝑛
Therefore,
𝑛

∑ 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥 𝑛 𝑦 𝑛 𝑚𝑎𝑑𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑒𝑟𝑚 𝑤𝑖𝑡ℎ 𝑥 𝑘 𝑦 𝑛−𝑘 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛
𝑘=0
𝑛

𝑛
𝑛
= ∑( )(
)
𝑘 𝑛−𝑘
𝑘=0

As both the methods are used to calculate the same value, we can equate both the values obtained in
Method 1 and Method 2.
𝑛

2𝑛
𝑛
𝑛
( ) = ∑ ( )(
)
𝑛
𝑘 𝑛−𝑘
𝑘=0

Hence Proved

2. Prove the identity
𝒏

𝒏
∑ 𝒌(𝒌 − 𝟏) ( ) = 𝟐𝒏−𝟐 𝒏(𝒏 − 𝟏)
𝒌

𝒌=𝟐

LHS
𝑛

𝑛...