Sum of n terms of series a b-a-1b-1-a-2b-2- -a-n-1b-n-1if a b-1-6 and a-b-1-3- is

The correct option is C s=ab+[ab+(a+b)+1]+[ab+2(a+b)+2^2]+....[ab+(n 1)(a+b)+(n 1)^2] =nab+(a+b)∑_r 1^n 1r+∑_r 1^n 1r^2 =nab+(a+b)n(n 1)/2+(n 1)(n)(2n 1)/6 =n/ ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Sum of n term...

Question

Sum of n terms of series
ab+(a+1)(b+1)+(a+2)(b+2)+....+(a+(n-1))(b+(n-1))
if $a b = \frac{1}{6} a n d (a + b) = \frac{1}{3}, i s$

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Solution

The correct option is C
$s = a b + [a b + (a + b) + 1] + [a b + 2 (a + b) + 2^{2}] + . . . . [a b + (n - 1) (a + b) + (n - 1)^{2}]$
$= n a b + (a + b) \sum_{r - 1}^{n - 1} r + \sum_{r - 1}^{n - 1} r^{2}$
$= n a b + (a + b) \frac{n (n - 1)}{2} + \frac{(n - 1) (n) (2 n - 1)}{6}$
$= \frac{n}{6} [1 + (n - 1) {1 + 2 n - 1}]$
$= \frac{n}{6} [1 + 2 n (n - 1)] = \frac{n}{6} (1 - 2 n + 2 n^{2})$

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Sum of n terms of series ab+(a+1)(b+1)+(a+2)(b+2)+....+(a+(n-1))(b+(n-1)) if ab=16 and(a+b)=13, is

The correct option is C s=ab+[ab+(a+b)+1]+[ab+2(a+b)+22]+....[ab+(n−1)(a+b)+(n−1)2] =nab+(a+b)∑n−1r−1r+∑n−1r−1r2 =nab+(a+b)n(n−1)2+(n−1)(n)(2n−1)6 =n6[1+(n−1){1+2n−1}] =n6[1+2n(n−1)]=n6(1−2n+2n2)

Sum of n terms of series
ab+(a+1)(b+1)+(a+2)(b+2)+....+(a+(n-1))(b+(n-1))
if $a b = \frac{1}{6} a n d (a + b) = \frac{1}{3}, i s$