If (1+x)ⁿ=_{r=0}ⁿ C_r x^r, then the value of C₀+ (C₀+C₁) + (C₀+C₁+C₂) +. . .+ (C₀+C₁+C₂++Cₙ) is [AP EAPCET 21 May 2025 S1]

Correct Answer is : (n+2)2n−1(n+2) 2ⁿ-1}(n+2)2n−1

Correct Answer is : (n+2)2n−1(n+2) 2ⁿ-1}(n+2)2n−1

Test Index

Binomial Theorem

Section: Mathematics

Question : 16 of 33

Marks: +1, -0

(1+x)^n=\sum\limits_{r=0}^{n} C_r x^r

C_0+ (C_0+C_1) + (C_0+C_1+C_2) +. . .+

(C_0+C_1+C_2+\dots+C_n)

Solution:

The given sum can be rewritten as

\sum\limits_{k=0}^{n} \sum\limits_{r=0}^{k} C_r

Now, changing order of summarion

\;\sum\limits_{r=0}^{n} \sum\limits_{k=r}^{n} C_r \;\; \left[ \because \sum\limits_{r=0}^{n} C_r=2^n\right]

\;\sum\limits_{r=0}^{n} C_r \sum\limits_{k=r}^{n} 1=\sum\limits_{r=0}^{n} C_r(n-r+1)

\;\Rightarrow (n+1) \sum\limits_{r=0}^{n} C_r-\sum\limits_{r=0}^{n} r \cdot C_r

\; \;\; =(n+1) 2^n-\sum\limits_{r=0}^{n} r \cdot{ }^{n-1} C_{r-1} \times \;\frac{n}{r}

\; \;\; =(n+1) 2^n-n \sum\limits_{r=0}^{n} n-1 C_r

\; \;\; =(n+1) 2^n-n \cdot 2^{n-1}

\; \;\; =n \cdot 2^n+2^n-n \cdot 2^{n-1}

\; \;\; =2^{n-1}(n+2)

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