1-3-3-5-5-7-2 n -12 n -1- n 4 n -6 n -1-3

1.3+3.5+5.7+… (2n 1)(2n+1) = n(4n+6n 1)/3 =∑ (2n 1)(2n+1) =∑ 4n^2 1 =4∑ n^2 ∑1 =4 (n)(n+1)(2n+1)/6 n =2/3n(n+1)(2n+1) n =n [ 2/3(n+1)(2n+1) 2 ] =n [ 2/3 ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

1.3+3.5+5.7+…...

Question

$1.3 + 3.5 + 5.7 + \dots (2 n - 1) (2 n + 1) = \frac{n (4 n + 6 n - 1)}{3}$

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Solution

$1.3 + 3.5 + 5.7 + \dots (2 n - 1) (2 n + 1) = \frac{n (4 n + 6 n - 1)}{3} = \sum (2 n - 1) (2 n + 1) = \sum 4 n^{2} - 1 = 4 \sum n^{2} - \sum 1 = 4 \frac{(n) (n + 1) (2 n + 1)}{6} - n = \frac{2}{3} n (n + 1) (2 n + 1) - n = n [\frac{2}{3} (n + 1) (2 n + 1) - 2] = n [\frac{2}{3} (2 n^{2} + 3 n + 1) - 1] = \frac{n}{3} [4 n^{2} + 6 n - 1]$

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Similar questions

$1.3 + 3.5 + 5.7 + \dots + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3} .$

Q. 1.3+3.5+5.7+....+

(2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

Q. Prove

1.3 + 3.5 + 5.7 + . . . + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

Prove that,

1.3+3.5+5.7+.............+(2n-1)(2n+1)=(n(4n²+6n-1))/3.

Q. 1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =

\frac{n (4 n^{2} + 6 n - 1)}{3}

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1.3+3.5+5.7+…(2n−1)(2n+1)=n(4n+6n−1)3

1.3+3.5+5.7+…(2n−1)(2n+1)=n(4n+6n−1)3=∑(2n−1)(2n+1)=∑4n2−1=4∑n2−∑1=4(n)(n+1)(2n+1)6−n=23n(n+1)(2n+1)−n=n[23(n+1)(2n+1)−2]=n[23(2n2+3n+1)−1]=n3[4n2+6n−1]

$1.3 + 3.5 + 5.7 + \dots (2 n - 1) (2 n + 1) = \frac{n (4 n + 6 n - 1)}{3}$