If sum_{r=1}^n(2r+1)=440,then n = ldots. | vedclass

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(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$. Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$. This is an arithmetic pro

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$20$

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(A) Given the sum $\sum_{r=1}^n(2r+1) = 440$. Expanding the summation,we get the series $3 + 5 + 7 + \ldots + (2n+1) = 440$. This is an arithmetic progression with first term $a = 3$,common difference $d = 2$,and number of terms $n$. The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n-1)d]$. Substituting the values: $\frac{n}{2}[2(3) + (n-1)(2)] = 440$. $\Rightarrow \frac{n}{2}[6 + 2n - 2] = 440$. $\Rightarrow \frac{n}{2}[2n + 4] = 440$. $\Rightarrow n(n + 2) = 440$. $\Rightarrow n^2 + 2n - 440 = 0$. Solving the quadratic equation: $(n + 22)(n - 20) = 0$. Since $n$ must be positive,$n = 20$.

If $\sum_{r=1}^n(2r+1)=440$,then $n = \ldots$.

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