Take a lazy tag $$$lazy_1$$$ in each node on the segement tree means that increase the value of $$$a_l$$$ by $$$lazy_1$$$ , $$$a_{l+1}$$$ by $$$2*lazy_1,\dots$$$ $$$a_{r}$$$ by $$$(r-l+1)*lazy_1$$$, and a lazy tag $$$lazy_2$$$ means that increase the value of $$$a_{l\dots r}$$$ by $$$lazy_2$$$. And maintain the sum.

When there is a query of type 1 and we'll divide the queried segment into two halves, increase the value of $$$lazy_1$$$ in the left node and right node by $$$1$$$. Then we increase the value of $$$lazy_2$$$ in the right node by $$$L_{rightson}-ql$$$ (here $$$L_{rightson}$$$ means left endpoint of right node,$$$ql$$$ is $$$l$$$ in the query).

When pushing down the lazy tags, you can push down $$$lazy_2$$$ as usual (I believe that you know.), and you should use summation formula of arithmetic sequence to push down $$$lazy_1$$$. concretely, we increase the sum we're pushing by $$$\frac{(lazy_1+lazy_1*(size))*size}{2}$$$ (here $$$size$$$ means the size of node we're pushing).

Forgive me for my poor expression.

void push_down_(int now,int l,int r,int k1,int k2){
	tr[now].sum+=(k1+k1*(r-l+1))*(r-l+1)/2+k2*(r-l+1);
	tr[now].lazy1+=k1;
	tr[now].lazy2+=k2;
}
void push_down(int now,int l,int r){
	int mid=(l+r)/2;
	push_down_(lson(now),l,mid,tr[now].lazy1,tr[now].lazy2);
	push_down_(rson(now),mid+1,r,tr[now].lazy1,tr[now].lazy2+tr[now].lazy1*(r-mid));
	tr[now].lazy1=tr[now].lazy2=0;
}
void update(int ql,int qr,int l,int r,int now,int k2){
	if(ql<=l&&r<=qr){
		tr[now].sum+=(1+r-l+1)*(r-l+1)/2+k2*(r-l+1);
		tr[now].lazy1+=1;
		tr[now].lazy2+=k2;
	}
	push_down(now,l,r);
	int mid=(l+r)/2;
	if(qr<=mid)update(ql,qr,l,mid,lson(now),k2);
	else if(ql>mid)update(ql,qr,mid+1,r,rson(now),k2);
	else update(ql,qr,l,mid,lson(now),k2),update(ql,qr,mid+1,r,rson(now),k2+mid+1-ql);
	push_up(now);
}