Question

Find the value of 6^2 +7^2 +8^2 +9^2 +......+50^2
(A) 42870
C) 42900
(B) 42925
(D) 1275​

Answer 1

Step-by-step explanation:

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$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve and verify the equation

$\frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{ 3} )$

$\huge\tt\underline\blue{Answer }$

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$⟹</p><p> \frac{1}{3} x - 4 = x - ( \frac{1}{2} + \frac{x}{3} )$

$⟹</p><p> \frac{x}{3} - 4 = x - ( \frac{3 + 2x}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - ( \frac{2x + 3}{6} )$

$⟹</p><p> \frac{x - 12}{3} = x - \frac{2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{6x - 2x - 3}{6}$

$⟹</p><p> \frac{x - 12}{3} = \frac{4x - 3}{6}$

cancelling 6( R.H.S) By 3 From L.H.S

$⟹ \frac{x - 12}{1} = \frac{4x - 3}{2} </p><p>$

$⟹</p><p>2(x - 12) = 4x - 3$

$⟹</p><p>2x - 24 = 4x - 3$

$⟹</p><p> - 24 + 3 = 4x - 2x$

$⟹</p><p> - 21 = 2x$

$⟹</p><p>x = - \frac{21}{2}$

CHECK:-

$⟹ \frac{ - \frac{21}{2} }{3} - 4 = - \frac{21}{2} - ( \frac{1}{2} + ( - ) \frac{ \frac{21}{2} }{3} )</p><p>$

$⟹</p><p> - \frac{21}{6} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{21}{6} )$

$⟹</p><p> - \frac{7}{2} - 4 = - \frac{21}{2} - ( \frac{1}{2} - \frac{7}{2} )$

$⟹</p><p> \frac{ - 7 - 8}{2} = - \frac{21}{2} - ( - \frac{6}{2} )$

$⟹ - \frac{15}{2} = - \frac{21}{2} - ( - 3) </p><p>$

$⟹</p><p> - \frac{15}{2} = - \frac{21}{2} + 3$

$⟹</p><p> - \frac{15}{2} = \frac{ - 21 + 6}{2} = - \frac{15}{2}$

THEREFORE,L.H.S=R.H.S

VERIFIED✔️

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HOPE IT HELPS YOU..

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Thankyou:)

Answer 2

Answer:

A) 42870

Step-by-step explanation:

Sum of squares of first n natural numbers

= n(n+1)(2n+1) / 6

6^2 + 7^2 + 8^2 + 9^2 +......+ 50^2

= (1^2 + 2^2 + 3^2+......+50^2) / 6

. - (1^2 + 2^2+ ....+5^2) / 6

= 50[50+1][2(50)+1] /6 -

5[5+1][2(5)+1] / 6

= 50(51)(101) / 6 - 5(6)(11) /6

= 257550 / 6 - 330 /6

= 42925 - 55

= 42870

THEREFORE 6^2 + 7^2 + 8^2 + 9^2 +.....50^2 = A) 42870 !