If 1 2 - a 1 - 2 2 - a 2 - 3 2 - a 3 - - n 2 - a n - 1 3 n n 2 - 1 - then find the value of a 7 -

We have, ( 1 ^ 2 a _ 1 ) +( 2 ^ 2 a _ 2 ) +......( n ^ 2 a _ n ) = 1 3 ( n ^ 2 1 ) .n For n=1 1 a _ 1 = 1 3 ( 1 1 ) =0 ⇒ a _ 1 ...

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Standard XII

Mathematics

Modulus of a Complex Number

If 1 2 - ...

Question

If $(1^{2} - a_{1}) + (2^{2} - a_{2}) + (3^{2} - a_{3}) + \dots (n^{2} - a_{n}) = \frac{1}{3} n (n^{2} - 1),$ then find the value of $a_{7} .$

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Solution

We have,
$(1^{2} - a_{1}) + (2^{2} - a_{2}) + . . . . . . (n^{2} - a_{n}) = \frac{1}{3} (n^{2} - 1) . n$

For $n = 1$
$1 - a_{1} = \frac{1}{3} (1 - 1) = 0$
$\Rightarrow a_{1} = 1$
$2^{2} - a_{2} + 0 = \frac{1}{3} (4 - 1) .2 = 2$

$∴$ $4 - a_{2} = 2 \Rightarrow a_{2} = 2$
$0 + 2 + (3^{2} - a_{3}) = \frac{1}{3} (9 - 1) .3 = 8$
$\Rightarrow 3^{2} - a_{3} = 6 \Rightarrow a_{3} = 3$
$8 + a (4^{2} - a_{4}) = \frac{1}{3} .4 (16 - 1) = 20$
$\Rightarrow 8 + 16 - a_{4} = 20 \Rightarrow a_{4} = 4$

Proceeding in similar manner $a_{7} = 7$

Hence, this is the answer.

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If (12−a1)+(22−a2)+(32−a3)+…(n2−an)=13n(n2−1), then find the value of a7.

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