1^2+2^2+3^2+4^2+........+50^2+2^2=?
Answers
Answer:
The sum of AP is -1275
Step-by-step explanation:
Given : AP 1^2-2^2+3^2-4^2+.....1
2
−2
2
+3
2
−4
2
+..... upto 50 terms.
To find : The sum of AP
Solution :
We can write the series as,
1^2-2^2+3^2-4^2+.....+49^2-50^21
2
−2
2
+3
2
−4
2
+.....+49
2
−50
2
(1^2+3^2+5^2+.....+49^2)-2^2(1^2+2^2+3^2+.....+25^2)(1
2
+3
2
+5
2
+.....+49
2
)−2
2
(1
2
+2
2
+3
2
+.....+25
2
)
Applying the series formula,
1^2+3^2+5^2+.....+49^2=\frac{n(2n+1)(2n-1)}{3}1
2
+3
2
+5
2
+.....+49
2
=
3
n(2n+1)(2n−1)
-2^2(1^2+2^2+3^2+.....+25^2=-2^2(\frac{n(n+1)(2n+1)}{6})−2
2
(1
2
+2
2
+3
2
+.....+25
2
=−2
2
(
6
n(n+1)(2n+1)
)
Substitute,
=(\frac{25(51)(49)}{3})-4\frac{25(26)(51)}{6})=(
3
25(51)(49)
)−4
6
25(26)(51)
)
=\frac{25(51)}{3}(49-52)=
3
25(51)
(49−52)
=\frac{25(51)}{3}(-3)=
3
25(51)
(−3)
=-1275=−1275
Therefore, The sum of AP is -1275.