Question

2²+3²+4²+5²+... up to 10 terms

Answer 1

Heya❣️❣️


first we know that

1sq+2sq+3sq+4sq.........=n(n+1) (2n+1)/6


here n=10 so sum of 10 terms =10(10+1)(2×10+1)/6
=10×11×21/6
=5×11×7
=385


as we want of 2 sq+3sq+4sq+....upto 10 terms so this=385-1=384.



Tháñkß ♥️

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\textsf{Let,} \\ \\ \sf \implies 2^2 \: + \: 3^2 \: + \: 4^2 \: + ...... + \: 11^2 \: = \: k$


$\textsf{Add 1^2 to both sides , } \\ \\ \sf \implies 1^2 \: + \: 2^2 \: + \: 3^2 \: +...... + 11^2 \: = \: k \: + \: {1}^{2}$



$\underline{\textsf{Using Formula : }} \\ \\ \sf \implies Sum \: of \: first \: n \: natural \: no.s \: = \: \dfrac{n(n \: + \: 1 )(2n \: + \: 1)}{6}$



$\sf \implies \dfrac{n(n \: + \: 1)(2n \: + \: 1 )}{6} \: = \: k \: + \: 1 \\ \\ \textsf{ Here , \: n \: = \: 11} \\ \\ \sf \implies \dfrac{11(11 \: + \: 1 )(2 \: \times \: 11 \: + \: 1)}{6} \: = \: k \: + \: 1 \\ \\ \sf \implies \dfrac{11 \: \times \: 12 \: \times \: (22 \: + \: 1)}{6} \: = \: k \: + \: 1 \\ \\ \sf \implies 11 \: \times \: 2 \: \times \: 23 \: = \: k \: + \: 1 \\ \\ \sf \implies 22 \: \times \: 23 \: = \: k \: + \: 1 \\ \\ \sf \implies 506\: = \: k \: + \: 1 \\ \\ \sf \implies 506\: - \: 1 \: = \: k \\ \\ \sf \: \: \therefore \: \: k \: = \: 505$


$\underline{\textsf{The required answer is 505.}}$